Baker’s Method and Tijdeman’s Argument

Abstract

This chapter is somewhat isolated and can be read (almost) independently of the others. We discuss in it the application of Baker’s method to Diophantine equations of Catalan type. We give a brief introduction to this method, show how it applies to classical Diophantine equations, and reproduce the beautiful argument of Tijdeman, who proved that Catalan’s equation has only finitely many solutions. Moreover, the solutions are bounded by an absolute effective constant (that is, a constant that can, in principle, be explicitly determined), which reduces the problem to a finite computation. Before the work of Mihăilescu this was the top achievement on Catalan’s problem.We also consider the more general equation of Pillai and show that it has finitely many solutions when one of the four variables is fixed.

Appendices

Appendix A: Number Fields

This is a brief synopsis of the algebraic number theory used in the present book. It cannot serve as introduction to algebraic number theory: its only purpose is to refresh the terminology and recall the basic facts. We give almost no proofs and almost no references, assuming that every reader will find the proofs of all (or, at least, most) of the statements from this appendix in his favorite algebraic number theory textbook(s).

A.1 Embeddings, Integral Bases, and Discriminant

An algebraic number field or, shorter, a number field is a finite extension of the field $\mathbb{Q}$ of rational numbers. Let K be a number field of degree $n = [K:\mathbb{Q}]$. Then there exist exactly n distinct embeddings $\sigma _{1},\ldots,\sigma _{n}: K\hookrightarrow \mathbb{C}$. An embedding σ is called real if $\sigma (K) \subset \mathbb{R}$ and complex otherwise. Complex embeddings enter the list $\sigma _{1},\ldots,\sigma _{n}$ in pairs: if σ is a complex embedding, then its complex conjugate $\bar{\sigma }$ is another complex embedding of K. In particular, the number of complex embeddings is always even.

A field with only real embeddings is called totally real, and a field with no real embeddings is called totally imaginary.

Denote by $\mathcal{O}_{K}$ the ring of algebraic integers from the number field K. The additive group of $\mathcal{O}_{K}$ is a free abelian group of rank n. Let $w_{1},\ldots,w_{n}$ be a $\mathbb{Z}$-basis of $\mathcal{O}_{K}$ (such a basis is usually called an integral basis of K). The quantity

$$\displaystyle{\mathcal{D}_{K} = \left (\det \left [\sigma _{i}(w_{j})\right ]_{1\leq i,j\leq n}\right )^{2}}$$

is independent of the choice of the integral basis $w_{1},\ldots,w_{n}$ and is called the discriminant of K. It is a nonzero rational integer.

Recall also Kronecker’s theorem: if $\alpha \in \mathcal{O}_{K}$ satisfies $\vert \sigma (\alpha )\vert \leq 1$ for any embedding σ of K, then either α = 0 or α is a root of unity. [It is proved in Appendix B; see Proposition B.2(2).]

A.2 Units, Regulator

The invertible elements of the ring $\mathcal{O}_{K}$ are called the units (or Dirichlet units) of K. The multiplicative group $\mathcal{U}_{K} = \mathcal{O}_{K}^{\times }$ of units is a finitely generated abelian group. More precisely, let t ₁ and 2t ₂ be the number of real and complex embeddings of K, respectively. Let Ω _K be the group of roots of unity from K (it is a finite cyclic group). The fundamental Dirichlet unit theorem asserts that $\mathcal{U}_{K}/\varOmega _{K}$ is a free abelian group of rank $r = t_{1} + t_{2} - 1$. (The inequality $r \leq t_{1} + t_{2} - 1$ is relatively easy to establish; the nontrivial part is $r \geq t_{1} + t_{2} - 1$.)

Let $\sigma _{1},\ldots,\sigma _{r+1}$ be a selection of embeddings of K, containing all real embeddings and one from each pair of complex embeddings ($r + 1 = t_{1} + t_{2}$), and let $\eta _{1},\ldots,\eta _{r}$ be a basis of the infinite part of $\mathcal{U}_{K}$ (usually called a system of fundamental units). Put e _i  = 1 if σ _i is real and e _i  = 2 if σ _i is complex, and consider the map $\mathcal{U}_{K} \rightarrow \mathbb{R}^{r}$ which associates to every $\eta \in \mathcal{U}_{K}$ the vector

$$\displaystyle{\left (e_{1}\log \left \vert \sigma _{1}(\eta )\right \vert,\ldots,e_{r}\log \left \vert \sigma _{r}(\eta )\right \vert \right ).}$$

The kernel of this map is Ω _K , and its image is a lattice in $\mathbb{R}^{r}$. The fundamental volume of this lattice, that is, the quantity

$$\displaystyle{\left \vert \det \left [e_{i}\log \left \vert \sigma _{i}(\eta _{j})\right \vert \right ]_{1\leq i,j\leq r}\right \vert }$$

(which is independent of the choice of the basic units $\eta _{1},\ldots,\eta _{r}$ and of our selection of the embeddings $\sigma _{1},\ldots,\sigma _{r+1}$), is called the regulator of K and is denoted by $\mathcal{R}_{K}$.

A.3 Ideals, Factorization

The ring $\mathcal{O}_{K}$ is a Dedekind ring; that is, it is Noetherian and integrally closed, and all its nonzero prime ideals are maximal. Hence it has the unique factorization property for ideals: every nonzero ideal of  $\mathcal{O}_{K}$ has a unique (up to the order) presentation as a product of prime ideals. In other words, the nonzero ideals of $\mathcal{O}_{K}$ form a free multiplicative semigroup, the (nonzero) prime ideals being its free generators.

A fractional ideal is a finitely generated $\mathcal{O}_{K}$-submodule of K. (Equivalently, an $\mathcal{O}_{K}$-module $\mathfrak{a} \subset K$ is a fractional ideal if there exists a nonzero α ∈ K such that $\alpha \mathfrak{a} \subset \mathcal{O}_{K}$.) For every nonzero fractional ideal $\mathfrak{a}$ there exists a unique fractional ideal $\mathfrak{a}^{-1}$ such that $\mathfrak{a}\mathfrak{a}^{-1} = (1)$. It follows that nonzero fractional ideals form a free abelian group, with prime ideals as free generators.

One often abuses the language, calling the fractional ideals as the ideals of K. Ideals of $\mathcal{O}_{K}$ are then referred to as integral ideals of K, and nonzero prime ideals of $\mathcal{O}_{K}$ are called the prime ideals of K (or just the primes of K). We follow this tradition in the present book.

If $\mathfrak{a}$ is an ideal of a number field K, and L a finite extension of K, then $\mathfrak{a}\mathcal{O}_{L}$ is an ideal of L, which is usually denoted by $\mathfrak{a}$ as well. (In case of confusion, one can specify which field is in mind, but this is seldom needed.) For instance, a prime ideal $\mathfrak{p}$ of K often becomes composite in L, and we can speak about the prime decomposition of $\mathfrak{p}$ in L.

A.4 Norm of an Ideal

Let $\mathfrak{a}$ be a nonzero ideal of $\mathcal{O}_{K}$. The norm of $\mathfrak{a}$, denoted by $\mathcal{N}\mathfrak{a}$, is, by definition, the cardinality of the residue ring $\mathcal{O}_{K}/\mathfrak{a}$. The norm of the zero ideal is, by definition, 0.

The norm is multiplicative: $\mathcal{N}(\mathfrak{a}\mathfrak{b}) = \mathcal{N}\mathfrak{a}\,\mathcal{N}\mathfrak{b}$. By multiplicativity, the norm function extends to all fractional ideals, defining a homomorphism from the group of (nonzero) fractional ideals to $\mathbb{Q}^{\times }$. This function is compatible with the usual norm map $K \rightarrow \mathbb{Q}$: the norm of the principal ideal (α) is equal to the absolute value of the norm of α.

If K is a Galois extension of $\mathbb{Q}$ with Galois group G, then for any ideal $\mathfrak{a}$ we have $\prod _{\sigma \in G}\mathfrak{a}^{\sigma } = (\mathcal{N}\mathfrak{a})$. (We write the Galois action by G exponentially.)

The norm defined above is called sometimes the absolute norm. More generally, let K be a number field, L a finite extension of K, and $\mathfrak{a}$ an ideal of L. Then one can define the L∕K-norm (called also relative norm) $\mathcal{N}_{L/K}\mathfrak{a}$, which will be an ideal of K. The L∕K-norm is multiplicative, and one also has the “transitivity relation”: if $K \subset L \subset M$ is a tower of number fields and $\mathfrak{a}$ is an ideal of M, then

$$\displaystyle{ \mathcal{N}_{L/K}\left (\mathcal{N}_{M/L}\mathfrak{a}\right ) = \mathcal{N}_{M/K}\mathfrak{a}. }$$

(A.1)

If L is a Galois extension of K with Galois group G and $\mathfrak{a}$ an ideal of L, then $\prod _{\sigma \in G}\mathfrak{a}^{\sigma } = \mathcal{N}_{L/K}\mathfrak{a}$.

If $\mathfrak{a}$ is an ideal of the number field K, then one can define the absolute norm $\mathcal{N}\mathfrak{a}$, which is a nonnegative rational number, and the $K/\mathbb{Q}$-norm $\mathcal{N}_{K/\mathbb{Q}}\mathfrak{a}$, which is an ideal of $\mathbb{Q}$. The two definitions agree in the sense that $\mathcal{N}_{K/\mathbb{Q}}\mathfrak{a}$ is the ideal generated by $\mathcal{N}\mathfrak{a}$.

A.5 Ideal Classes, the Class Group

The multiplicative group of fractional ideals has a subgroup consisting of principal ideals (it is isomorphic to $K^{\times }/\mathcal{O}_{K}^{\times }$). The corresponding quotient group is called the group of classes of ideals or, shorter, the class group of K. The class group is denoted by $\mathcal{H}_{K}$; its elements are called classes of ideals, or ideal classes, or simply classes.

It is fundamental and nontrivial that the class group is finite. Its cardinality is called the class number of K; it is denoted by h _K . Together with the discriminant $\mathcal{D}_{K}$ and regulator $\mathcal{R}_{K}$, the class number belongs to the most important numerical invariants of the field K.

A.6 Prime Ideals, Ramification

Any prime ideal of K contains exactly one prime number, called the underlying prime of this prime ideal. Let $\mathfrak{p}$ be a prime ideal and $p$ its underlying prime. Then $\mathcal{O}_{K}/\mathfrak{p}$ is the finite field $\mathbb{F}_{p^{f}}$, where the integer f satisfies $f \leq n = [K:\mathbb{Q}]$ and is called the residue field degree of $\mathfrak{p}$ (over $\mathbb{Q}$).

Further, the positive integer e such that $\mathfrak{p}^{e}$ divides p, but $\mathfrak{p}^{e+1}$ does not, is called the ramification index of $\mathfrak{p}$ (over $\mathbb{Q}$). The ideal $\mathfrak{p}$ is ramified (over $\mathbb{Q}$) if e > 1 and unramified otherwise. The product ef is called the local degree, or simply the degree of $\mathfrak{p}$ (over $\mathbb{Q}$).

Conversely, let p be a prime number. Then only finitely many (actually, no more than n) prime ideals of K lie above p. Let $\mathfrak{p}_{1},\ldots,\mathfrak{p}_{s}$ be these prime ideals, $e_{1},\ldots,e_{s}$ their ramification indices, and $f_{1},\ldots,f_{s}$ their residue field degrees, respectively. Then we have the factorization $(p) = \mathfrak{p}_{1}^{e_{1}}\cdots \mathfrak{p}_{s}^{e_{s}}$. Comparing the absolute norms of both parts, we obtain the basic identity $e_{1}f_{1} + \cdots + e_{s}f_{s} = n$ (“global degree is the sum of local degrees”).

We say that p splits completely in K if p decomposes in K into a product of n distinct prime ideals; equivalently, p splits completely if all prime ideals of K above p are of local degree 1.

We say that the prime p is ramified in K if at least one of the ideals $\mathfrak{p}_{1},\ldots,\mathfrak{p}_{s}$ is ramified over $\mathbb{Q}$ , that is, if at least one of the numbers $e_{1},\ldots,e_{s}$ is greater than 1. Otherwise, the prime p is unramified in K. The prime p is totally ramified in K if $s = 1,e_{1} = n,f_{1} = 1$, that is, if $(p) = \mathfrak{p}^{n}$ for some prime ideal $\mathfrak{p}$. In this case, p is also totally ramified in any subfield of K (distinct from $\mathbb{Q}$).

One proves the existence of an integral ideal $\mathfrak{d}_{K}$, called the different of K, with the following properties: $\mathcal{N}(\mathfrak{d}_{K}) = \left \vert \mathcal{D}_{K}\right \vert$, and a prime ideal $\mathfrak{p}$ is ramified if and only if $\mathfrak{p}\mid \mathfrak{d}_{K}$. Consequently, a prime number p is ramified in K if and only if it divides the discriminant  $\mathcal{D}_{K}$. In particular, there are only finitely many ramified primes.

Minkowski’s theorem asserts that $\vert \mathcal{D}_{K}\vert > 1$ for $K\neq \mathbb{Q}$. Equivalently, if $K\neq \mathbb{Q}$ then at least one prime number is ramified in K.

More generally, let K be a number field, L a finite extension of K, and $\mathfrak{P}$ a prime ideal of L. Then $\mathfrak{p} = \mathfrak{P} \cap \mathcal{O}_{K}$ is a prime ideal of K, called the underlying prime ideal of  $\mathfrak{P}$ in K. One can define now the ramification index and the residue field degree of $\mathfrak{P}$ over K, the relative different $\mathfrak{d}_{L/K}$, the relative discriminant $\mathcal{D}_{L/K} = \mathcal{N}_{L/K}(\mathfrak{d}_{L/K})$, and so on. We shall use the following properties of the different:

• Differents are multiplicative in towers; that is, if $K \subseteq L \subseteq M$ is a tower of number fields, then $\mathfrak{d}_{M/K} = \mathfrak{d}_{M/L}\mathfrak{d}_{L/K}$.

• If $\alpha \in \mathcal{O}_{L}$ is such that L = K(α) and $f(x) \in \mathcal{O}_{K}[x]$ is the minimal polynomial of α over K then $\mathfrak{d}_{L/K}$ divides f′(α).

A.7 Galois Extensions

Let K be a finite Galois extension of $\mathbb{Q}$ with the Galois group $G = \mathrm{Gal}(K/\mathbb{Q})$. Let $p$ be a prime number and $\mathfrak{p}$ a prime ideal of K above p . Then for any σ ∈ G the prime ideal $\mathfrak{p}^{\sigma }$ again lies above p . Thus, G acts in the obvious way on the set $\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{s}\}$ of the prime ideals of $K$ above p. It is important that this action is transitive, which implies that the ramification indices $e_{1},\ldots,e_{s}$ are all equal, and the same is true for the residue field degrees $f_{1},\ldots,f_{s}$. One calls the number $e = e_{1} = \cdots = e_{s}$ the ramification index of p in K; similarly, $f = f_{1} = \cdots = f_{s}$ is called the residue field degree of p in K.

Again, let $\mathfrak{p}$ be a prime ideal of K and p the underlying prime number. The decomposition group of $\mathfrak{p}$ is the subgroup of G stabilizing $\mathfrak{p}$:

$$\displaystyle{G_{\mathfrak{p}} =\{\sigma \in G: \mathfrak{p}^{\sigma } = \mathfrak{p}\}.}$$

The inertia group is the subgroup of $G_{\mathfrak{p}}$ defined by

$$\displaystyle{I_{\mathfrak{p}} =\{\sigma \in G_{\mathfrak{p}}: \mbox{ $\alpha ^{\sigma } \equiv \alpha \mathrm{mod}\,\mathfrak{p}$ for all $\alpha \in \mathcal{O}_{K}$}\}.}$$

We have $\vert G_{\mathfrak{p}}\vert = ef$ and $\vert I_{\mathfrak{p}}\vert = e$. The inertia group is a normal subgroup of the decomposition group, the quotient $G_{\mathfrak{p}}/I_{\mathfrak{p}}$ being canonically isomorphic to the Galois group of the residue field $\mathcal{O}_{K}/\mathfrak{p}$ over $\mathbb{Z}/p\mathbb{Z}$. In particular, it is a cyclic group of f elements.

If $\mathfrak{p}$ is unramified then $I_{\mathfrak{p}}$ is trivial and $G_{\mathfrak{p}}$ is itself a cyclic group of f elements. Moreover, it has a canonical generator $\varphi =\varphi _{\mathfrak{p}} \in G_{\mathfrak{p}}$, called the Frobenius element. The latter is defined as the unique element of $G_{\mathfrak{p}}$ such that $\alpha ^{\varphi } \equiv \alpha ^{p}\ \mathrm{mod}\,\mathfrak{p}$ for all $\alpha \in \mathcal{O}_{K}$. Thus, to every unramified prime ideal $\mathfrak{p}$ of K, we associate its Frobenius element $\varphi _{\mathfrak{p}}$.

One has similar terminology and statements in the relative case, for the Galois extensions L∕K. We shall be very brief. Let $\mathfrak{P}$ be a prime ideal of $L$ and $\mathfrak{p}$ the underlying prime ideal of K. The subgroup $G_{\mathfrak{P}/\mathfrak{p}}$ of $G = \mathrm{Gal}(L/K)$ preserving $\mathfrak{P}$ is called the decomposition group of  $\mathfrak{P}$ over K. If $\mathfrak{P}$ is unramified over K then $G_{\mathfrak{P}/\mathfrak{p}}$ is a cyclic group generated by the Frobenius element $\varphi =\varphi _{\mathfrak{P}/\mathfrak{p}}$, which is the unique element of $G_{\mathfrak{P}/\mathfrak{p}}$ satisfying $\alpha ^{\varphi } \equiv \alpha ^{\mathcal{N}\mathfrak{p}}\ \mathrm{mod}\,\mathfrak{P}$ for all $\alpha \in \mathcal{O}_{L}$.

A.8 Valuations

Let K be a field (not just a number field). A valuation v on K is a real-valued function x↦ | x |  _v on K with the following properties:

• 1. We have | α |  _v  ≥ 0 for all α ∈ K; also, | α |  _v  = 0 if and only if α = 0.

• 2. For any α, β ∈ K we have $\vert \alpha \beta \vert _{v} = \vert \alpha \vert _{v}\vert \beta \vert _{v}$.

• 3. There exists a constant c _v  > 0 such that for any α, β ∈ K we have

  $$\displaystyle{ \vert \alpha +\beta \vert _{v} \leq c_{v}\max \left \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\right \}\,. }$$

  (A.2)

The valuation, defined by | 0 |  _v  = 0 and | α |  _v  = 1 for any α ≠ 0, is called trivial. In the sequel, trivial valuations are excluded from consideration; thus, from now on, when we say valuation, we mean nontrivial valuation.

If v is a valuation and a a positive real number then the function $x\mapsto \vert x\vert _{v}^{a}$ is again a valuation (with the constant $c_{v}^{a}$ instead of c _v ). Two valuations v and v′ are called equivalent if there exists a > 0 such that $\vert x\vert _{v'} = \vert x\vert _{v}^{a}$ for any x ∈ K.

A valuation v is called non-Archimedean if (A.2) holds with c _v  = 1 and Archimedean otherwise. If v is (non-)Archimedean, then all equivalent valuations are (non-)Archimedean as well.

One usually denotes by M _K the set of all classes of equivalent valuations of the field $K$ (these classes are called places). It is practical, however, to choose a representative in each class (“normalize the valuations”) and to denote by M _K the set of chosen representatives.

In general, there is no “canonical” way to normalize the valuations of a number field, and, except the case $K = \mathbb{Q}$, there is even no commonly accepted normalization: different normalizations are used in different sources. Below we describe the normalizations adopted in this book.

First of all, recall the commonly used normalization for $K = \mathbb{Q}$. We have

$$\displaystyle{M_{\mathbb{Q}} =\{ \text{the prime numbers}\} \cup \{\infty \},}$$

where the symbol ∞ corresponds to the usual absolute value: | α |  _∞  =  | α | for $\alpha \in \mathbb{Q}$, and each prime number p corresponds to the standard p-adic valuation, defined as follows. Let α be a nonzero rational number. We define the p-adic order Ord _p α as the unique integer m such that α = p ^m a∕b with $a,b \in \mathbb{Z}$ satisfying gcd(p , ab) = 1. Next, we define $\vert \alpha \vert _{p} = p^{-\mathrm{Ord}_{p}(\alpha )}$. By definition, one puts Ord _p 0 = +∞ and | 0 |  _p  = 0. With this normalization, for any nonzero $\alpha \in \mathbb{Q}$, one has the product formula $\prod _{v\in M_{\mathbb{Q}}}\vert \alpha \vert _{v} = 1$.

Now let K be an arbitrary number field. Then

$$\displaystyle\begin{array}{rcl} M_{K}& =& \{\mbox{ the prime ideals of $K$}\} \cup \{\mbox{ the real embeddings of $K$}\} \cup {}\\ & &\{\mbox{ the pairs of complex conjugate embeddings of $K$}\}. {}\\ \end{array}$$

If v ∈ M _K is a real embedding σ, then, for α ∈ K, we define | α |  _v  =  | σ(α) | . If v is a pair of complex conjugate embeddings $\sigma,\bar{\sigma }$, then we define $\vert \alpha \vert _{v} = \vert \sigma (\alpha )\vert ^{2}$. Finally, if v is a prime ideal $\mathfrak{p}$ then $\vert \alpha \vert _{v} = (\mathcal{N}\mathfrak{p})^{-\mathrm{Ord}_{\mathfrak{p}}(\alpha )}$ (the $\mathfrak{p}$ -adic valuation). Here $\mathrm{Ord}_{\mathfrak{p}}(\alpha )$ is the $\mathfrak{p}$-adic order of α, defined by $(\alpha ) = \mathfrak{p}^{\mathrm{Ord}_{\mathfrak{p}}(\alpha )}\mathfrak{a}\mathfrak{b}^{-1}$, where $\mathfrak{a},\mathfrak{b}$ are integral ideals of K coprime with $\mathfrak{p}$ (with the convention $\mathrm{Ord}_{\mathfrak{p}}(0) = +\infty$). With this normalization we again have the product formula

$$\displaystyle{\prod _{v\in M_{K}}\vert \alpha \vert _{v} = 1\qquad (\alpha \in K^{\times })\,.}$$

The valuations (and places) corresponding to the embeddings of K are called infinite; they are Archimedean. The $\mathfrak{p}$-adic valuations are called finite; they are non-Archimedean.

Let K be a number field and L a finite extension of K. We say that w ∈ M _L lies above v ∈ M _K and v lies below w (notation: w∣v) if the restriction of w to K is equivalent to v. There are only finitely many w ∈ M _L above a given v ∈ M _K . With our normalization, for any α ∈ L, we have the identity

$$\displaystyle{\left \vert \mathcal{N}_{L/K}\alpha \right \vert _{v} =\prod _{w\mid v}\vert \alpha \vert _{w}.}$$

In particular, if α ∈ K then

$$\displaystyle{ \prod _{w\mid v}\vert \alpha \vert _{w} = \vert \alpha \vert _{v}^{[L:K]}. }$$

(A.3)

Now let τ: K → K′ be an isomorphism of number fields. Then τ induces a natural map $\tau _{{\ast}}: M_{K} \rightarrow M_{K'}$. Indeed, for a v ∈ M _K , we define a valuation τ _∗ v of K′ by $\vert \beta \vert _{\tau _{{\ast}}v} = \vert \tau ^{-1}(\beta )\vert _{v}$. A priori, τ _∗ v is merely equivalent to a valuation from M _K′; however, it is easy to show that it is “correctly” normalized and thereby belongs to M _K′.

Indeed, if $v$ is $\mathfrak{p}$-adic, then τ _∗ v is $\mathfrak{p}'$-adic, where $\mathfrak{p}' =\tau (\mathfrak{p})$. It remains to notice that prime ideals $\mathfrak{p}$ of K and $\mathfrak{p}'$ of K′ have the same norm. If v is infinite and corresponds to the embedding σ of K, then τ _∗ v corresponds to the embedding σ ∘τ of K′, and if σ is real (respectively, complex), then σ ∘τ is real (respectively, complex) as well.

Recall in conclusion the following simple property of non-Archimedean valuations, which is widely exploited in the present book. It will be more convenient to use the notion of $\mathfrak{p}$-adic order rather than valuation.

Lemma A.1.

Let K be a number field, $\alpha _{0},\ldots,\alpha _{m} \in K$ , and  $\mathfrak{p}$ a prime ideal of K. Assume that α ₀ ≠ 0 and that $\mathrm{Ord}_{\mathfrak{p}}(\alpha _{0}) < \mathrm{Ord}_{\mathfrak{p}}(\alpha _{k})$ for k = 1,…,m. Then

$$\displaystyle{\mathrm{Ord}_{\mathfrak{p}}(\alpha _{0} + \cdots +\alpha _{m}) = \mathrm{Ord}_{\mathfrak{p}}(\alpha _{0})\,.}$$

In particular, $\alpha _{0} + \cdots +\alpha _{m}\neq 0$ .

A.9 Dedekind ζ-Function

The Riemann ζ-function is defined, for s > 1, by

$$\displaystyle{ \zeta (s) =\sum _{ n=1}^{\infty }n^{-s}. }$$

(A.4)

The arithmetical significance of this function comes from Euler’s observation that it decomposes into the following infinite product:

$$\displaystyle{ \zeta (s) =\prod _{p}\left (1 - p^{-s}\right )^{-1}, }$$

(A.5)

where p runs over all prime numbers (the Euler product formula).

Since the series $\sum _{n=1}^{\infty }n^{-1}$ diverges to infinity, we have ζ(s) → +∞ as s tends to 1 from the right. Using the obvious inequality

$$\displaystyle{(n + 1)^{-s} \leq \int _{ n}^{n+1}x^{-s}dx \leq n^{-s},}$$

we obtain

$$\displaystyle{\int _{1}^{+\infty }x^{-s}dx \leq \zeta (s) \leq 1 +\int _{ 1}^{+\infty }x^{-s}dx.}$$

Since $\int _{1}^{+\infty }x^{-s}dx = (s - 1)^{-1}$, we find the exact asymptotic behavior of ζ(s) as s approaches 1 from the right:

$$\displaystyle{\zeta (s) = (s - 1)^{-1} + O(1).}$$

In particular,

$$\displaystyle{ \lim _{s\downarrow 1}(s - 1)\zeta (s) = 1. }$$

(A.6)

The Dedekind ζ-function is a proper generalization of the Riemann ζ-function for the needs of algebraic number theory. Given a number field K, we define the Dedekind ζ-function ζ _K (s) by

$$\displaystyle{\zeta _{K}(s) =\sum _{\mathfrak{a}}\mathcal{N}(\mathfrak{a})^{-s},}$$

the sum being over all nonzero integral ideals of K. Obviously, $\zeta _{\mathbb{Q}}(s) =\zeta (s)$. Again, the series converges for s > 1, and we have the Euler product formula

$$\displaystyle{ \zeta _{K}(s) =\prod _{\mathfrak{p}}\left (1 -\mathcal{N}(\mathfrak{p})^{-s}\right )^{-1}, }$$

(A.7)

the product being over the prime ideals of K. One also has an analogue for the “residue formula” (A.6):

$$\displaystyle{ \lim _{s\downarrow 1}(s - 1)\zeta _{K}(s) = \frac{2^{t_{1}}(2\pi )^{t_{2}}\mathcal{R}_{K}h_{K}} {\omega \sqrt{\vert \mathcal{D}_{K } \vert }}, }$$

(A.8)

where t ₁ and 2t ₂ are the numbers of real and of complex embeddings and ω is the number of roots of unity in K. The proof again relies on approximating certain sums by integrals, but it is much more involved.

The series (A.4) converges not only for real s > 1 but also for complex s with Re s > 1, and the sum is analytic on the half-plane Re s > 1. Riemann showed that it satisfies a functional equation, which allows one to continue ζ(s) to a meromorphic function on $\mathbb{C}$ with a single pole at s = 1. The same is true for the Dedekind ζ-function ζ _K (s). The complex analytic theory of the ζ-function is deep and important, but we do not use it in this book.

A.10 Chebotarev Density Theorem

One uses the ζ-function to establish various “density theorems” about prime numbers and prime ideals. The most important of them is the Chebotarev density theorem.

First of all, we define the notion of Dirichlet density. Combining the Euler product formula (A.5) and the residue formula (A.6), we obtain

$$\displaystyle{ \sum _{p} -\log (1 - p^{-s}) =\log \frac{1} {s - 1} + O(1) }$$

(A.9)

for s > 1. Using that −log(1 − z) = z + O( | z | ²) for | z | ≤ 1∕2, we may write the left-hand side of (A.9) as

$$\displaystyle{\sum _{p}p^{-s} + O\left (\sum _{ p}p^{-2s}\right ).}$$

The O(⋅ )-term is obviously bounded for s > 1, and we obtain

$$\displaystyle{ \sum _{p}p^{-s} =\log \frac{1} {s - 1} + O(1). }$$

(A.10)

Now let A be a set of prime numbers. We say that the set A is regular if there exists a real number a such that

$$\displaystyle{\sum _{p\in A}p^{-s} = a\log \frac{1} {s - 1} + O(1)}$$

for s > 1. The number a is called the Dirichlet density of the set A.

Now we are ready to state the theorem of Chebotarev. Let K be a finite Galois extension of $\mathbb{Q}$ and let G be its Galois group. Recall that to every prime ideal $\mathfrak{p}$ of K, unramified over $\mathbb{Q}$, we associated its Frobenius element $\varphi _{\mathfrak{p}}$ (see Sect. A.7). Now, in the opposite direction, to every prime number p, unramified in K, we associate the subset of G, consisting of the Frobenius elements of all prime ideals above p. This subset is a full conjugacy class⁴ of G; it is called the Artin symbol of p (with respect to K) and is denoted by $\left [ \frac{p} {K}\right ]$.

Chebotarev proved that conversely, every element from G serves as the Frobenius element for infinitely many prime ideals of K, and, consequently, each conjugacy class serves as the Artin symbol for infinitely many prime numbers. In particular, the Artin map

$$\displaystyle\begin{array}{rcl} \{\mbox{ prime numbers unramified in $K$}\}& \rightarrow \{\mbox{ conjugacy classes of $G$}\}& {}\\ p& \mapsto \left [ \frac{p} {K}\right ] & {}\\ \end{array}$$

is surjective. In fact, Chebotarev proved much more than this: given a conjugacy class S of G, the set of prime numbers with Artin symbol S is regular and has the “correct” Dirichlet density $\frac{\vert S\vert } {\vert G\vert }$.

In the case when K is a cyclotomic field, Chebotarev’s theorem becomes the classical theorem of Dirichlet about primes in arithmetical progressions (see Sect. 5.2.2).

A similar statement holds in the relative case. Let K be a number field. Euler’s product (A.7) together with the residue formula (A.8) implies that

$$\displaystyle{\sum _{\mathfrak{p}}\mathcal{N}(\mathfrak{p})^{-s} =\log \frac{1} {s - 1} + O(1),}$$

where the sum extends to all the prime ideals of K. We say that a set A of prime ideals is regular of Dirichlet density a if

$$\displaystyle{\sum _{\mathfrak{p}\in A}\mathcal{N}(\mathfrak{p})^{-s} = a\log \frac{1} {s - 1} + O(1)}$$

for s > 1.

One simple remark would be relevant here: the set of prime ideals of degree 1 (over $\mathbb{Q}$) is regular and of Dirichlet density 1. Indeed, since there is at most $n = [K: \mathbb{Q}]$ prime ideals over every rational prime, we have, for s > 1,

$$\displaystyle{\sum _{f_{\mathfrak{p}}\geq 2}\mathcal{N}(\mathfrak{p})^{-s} =\sum _{ p}\sum _{{ \mathfrak{p}\mid p\above 0.0pt f_{ \mathfrak{p}}\geq 2} }p^{-f_{\mathfrak{p}}s} \leq n\sum _{ p}p^{-2s} = O(1).}$$

Hence, the prime ideals of degree at least 2 form a regular set of Dirichlet density 0, and those of degree 1 form a regular set of density 1.

Now let L∕K be a finite Galois extension with Galois group G, and let $\mathfrak{p}$ be a prime ideal of $K$ unramified in L. Then the Artin symbol of $\mathfrak{p}$ (with respect to the extension L∕K) is

$$\displaystyle{\left [ \frac{\mathfrak{p}} {L/K}\right ] =\{\varphi _{\mathfrak{P}/\mathfrak{p}}: \mathfrak{P}\vert \mathfrak{p}\};}$$

it is a full conjugacy class of G. The Chebotarev density theorem asserts that the set of prime ideals whose Artin symbol is a given conjugacy class S is regular and of Dirichlet density $\frac{\vert S\vert } {\vert G\vert }$. In particular, the Artin map

$$\displaystyle\begin{array}{rcl} \left \{\mbox{ prime ideals of $K$ unramified in $L$}\right \}& \rightarrow & \{\mbox{ conjugacy classes of $G$}\} \\ \mathfrak{p}& \mapsto & \left [ \frac{\mathfrak{p}} {L/K}\right ] {}\end{array}$$

(A.11)

is surjective.

Since the prime ideals of degree 1 have Dirichlet density 1, every conjugacy class of G serves as the Artin symbol for infinitely many prime ideals of K of degree 1.

A.11 Hilbert Class Field

According to the theorem of Minkowski, mentioned in Sect. A.6, for every nontrivial extension of $\mathbb{Q}$, there exists a prime number ramified in this extension. This theorem, however, does not extend to arbitrary number fields: a number field K may have, in general, nontrivial extensions where no prime ideal of K ramifies. For example, so is the extension $\mathbb{Q}(\sqrt{-5},\sqrt{-1})$ of the field $\mathbb{Q}(\sqrt{-5})$. Such extensions are called unramified at finite places, because prime ideals correspond to the finite places of the field K (see Sect. A.8). We want to extend this to infinite places as well.

There are two types of infinite places: those corresponding to real embeddings of K (real infinite places) and those corresponding to pairs of complex conjugate embeddings (complex infinite places).

Let L∕K be a finite extension of number fields. We say that an infinite place w of L is ramified over K if w is a complex place, but the underlying place v of K is real. If both v and w are real, or both are complex, we say that w is unramified over K. An infinite place v of K is unramified in L if all the places above v are unramified over K. This happens if either v is complex or v is real and so are all the places above.

We say that L is unramified over K at infinity if every infinite place of K is unramified in L. Equivalently, L is unramified over K at infinity if every real embedding of K extends only to real embeddings of L. In particular, if K is totally real then only totally real extensions K are unramified at infinity, and if K is totally imaginary, then any extension of K is unramified at infinity.

Finally, we say that L is unramified over K if it is unramified at all places, both finite and infinite. For example, the already mentioned field $\mathbb{Q}(\sqrt{-5},\sqrt{-1})$ is an unramified extension of $\mathbb{Q}(\sqrt{-5})$.

Now let L be a finite abelian extension of a number field K. This means that L∕K is a Galois extension and the Galois group $G = \mathrm{Gal}(K/\mathbb{Q})$ is abelian.

Let $\mathfrak{p}$ be a prime ideal of K unramified in L. In Sect. A.10 we associated to it the Artin symbol $\left [ \frac{\mathfrak{p}} {L/K}\right ]$, which is a conjugate class of G. Since G is abelian, this class consists of a single element. Thus, for the abelian extensions, the Artin map has values in G itself rather than in the set of conjugacy classes.

If, in addition, we assume that the abelian extension L∕K is unramified at all finite places, then we have the map

$$\displaystyle{\{\mbox{ prime ideals of $K$}\} \rightarrow G.}$$

Extending it by linearity, we obtain a group homomorphism I → G, where I is the group of all ideals of the field K. It is called the Artin homomorphism of the abelian extension L∕K. Chebotarev density theorem implies that the Artin homomorphism is surjective. Describing its kernel is the central problem of the Class Field Theory. The principal result (usually referred to as Artin’s reciprocity law) states: if the abelian extension L∕K is unramified (everywhere, at all finite and infinite places), then the kernel of the Artin map contains the group of principal ideals.⁵ This defines a surjective homomorphism $\mathcal{H}_{K} \rightarrow G$ (also called the Artin map or Artin homomorphism), where $\mathcal{H}_{K}$ is the ideal class group of K.

Conversely, let K be a number field, and fix a surjective homomorphism $\mathcal{H}_{K} \rightarrow G$ of the class group of K onto a finite abelian group G. Then there exists a unique (in the given algebraic closure) unramified abelian extension L∕K with Galois group G and such that the fixed homomorphism is the Artin homomorphism of L∕K. In particular, there exists a unique unramified abelian extension of K with Galois group canonically isomorphic to  $\mathcal{H}_{K}$, the isomorphism being given by the Artin map. It is called the Hilbert class field of K.

Certain analogues of the results described above hold for ramified abelian extensions as well. A very good account (without proofs) of the Class Field Theory can be found in an appendix to Washington’s book [136]. Among the complete (with all proofs) expositions of the Class Field Theory, the most systematic one is, probably, due to Neukirch [104, 105]. The recent book of Childress [24] is a good introductory text for an unsophisticated reader. One may also consult the famous books of Artin and Tate [2], Cassels and Frölich [19], Weil [137] , and Lang [61].

Appendix B: Heights

In this appendix we introduce the height function on the field of algebraic numbers $\bar{\mathbb{Q}}$. We start with an informal discussion, which should motivate the definition.

Intuitively, the height should measure the “size” of an algebraic number. We expect our height function to have the following properties:

• The height of an algebraic number is a nonnegative real number.

• The height should “behave well” with respect to algebraic operations. (That is, one should have easy and efficient upper estimates for the heights of $\alpha +\beta$ and α β in terms of the heights (and, perhaps, degrees) of α and β.)

• For a given C > 0, there should exist only finitely many algebraic numbers of degree and height bounded by C.

Of course, one can imagine many functions with these properties. Our purpose is to define the one which is the most convenient to use.

On integers, the usual absolute value is an adequate measure of size. Thus, we define the height of a nonzero $\alpha \in \mathbb{Z}$ by H(α) =  | α | and H(0) = 1, the latter exception being made just for compatibility with the more general definition below.

On rational numbers, the absolute value is no longer adequate: there exist infinitely many rational numbers of bounded absolute value. To obtain finiteness, one should bound both the numerator and the denominator. Thus, for $\alpha = a/b \in \mathbb{Q}$, where $a,b \in \mathbb{Z}$ and gcd(a, b) = 1, we define H(α) = max{ | a | , | b | }.

Next, we wish to extend this definition to all algebraic numbers. One idea is to observe that bX − a is the minimal polynomial of the rational number a∕b over $\mathbb{Q}$. Hence, for $\alpha \in \bar{ \mathbb{Q}}$, we may put $\mathrm{H}(\alpha ) =\max \left \{\vert a_{0}\vert,\ldots,\vert a_{n}\vert \right \}$, where $a_{n}X^{n} + \cdots + a_{0}$ is the minimal polynomial of α over $\mathbb{Z}$. Indeed, one extensively used this definition of height in the past. However, it is rather inconvenient from many points of view, and it was eventually abandoned in favor of a different definition, due to A. Weil.

The definition of Weil’s height is motivated by the following observation: the height of a rational number α = a∕b, originally defined as max{ | a | , | b | }, satisfies the identity

$$\displaystyle{ \mathrm{H}(\alpha ) =\prod _{v\in M_{\mathbb{Q}}}\max \{1,\vert \alpha \vert _{v}\}. }$$

(B.1)

(The proof is an easy exercise.) Weil’s height is the logarithm⁶ of the right-hand side of this identity, properly generalized to number fields.

Now we are ready to give a formal definition of the height. Let K be a number field. Define the logarithmic K-height (or, simply, the K-height) $\mathrm{h}_{K}: K \rightarrow \mathbb{R}$ by

$$\displaystyle{\mathrm{h}_{K}(\alpha ) =\sum _{v\in M_{K}}\log \max \{1,\vert \alpha \vert _{v}\}.}$$

In the sequel, we use the notation⁷

$$\displaystyle{\|\alpha \|_{v} =\max \{ 1,\vert \alpha \vert _{v}\}\,,}$$

so that

$$\displaystyle{\mathrm{h}_{K}(\alpha ) =\sum _{v\in M_{K}}\log \|\alpha \|_{v}\,.}$$

Proposition B.1.

If K and L are two number fields, then for any $\alpha \in K \cap L$ we have

$$\displaystyle{ \frac{\mathrm{h}_{K}(\alpha )} {[K:\mathbb{Q}]} = \frac{\mathrm{h}_{L}(\alpha )} {[L:\mathbb{Q}]}\,. }$$

(B.2)

Proof.

Assume first that K ⊂ L. As follows from (A.3), for any α ∈ K and v ∈ M _K , we have

$$\displaystyle{[L:K]\log \|\alpha \|_{v} =\sum _{w\mid v}\log \|\alpha \|_{w}\,,}$$

where the sum extends to all w ∈ M _L lying above this v. Hence for any α ∈ K we have

$$\displaystyle\begin{array}{rcl} \frac{\mathrm{h}_{L}(\alpha )} {[L:\mathbb{Q}]}& =& \frac{1} {[L:\mathbb{Q}]}\sum _{w\in M_{L}}\log \|\alpha \|_{w} = \frac{1} {[L:\mathbb{Q}]}\sum _{v\in M_{K}}\sum _{w\mid v}\log \|\alpha \|_{w} {}\\ & =& \frac{[L:K]} {[L:\mathbb{Q}]} \sum _{v\in M_{K}}\log \|\alpha \|_{v} = \frac{\mathrm{h}_{K}(\alpha )} {[K:\mathbb{Q}]}\,, {}\\ \end{array}$$

which proves (B.2) in the case K ⊂ L. In the general case, let M be a number field containing both K and L. Then for any $\alpha \in K \cap L$

$$\displaystyle{ \frac{\mathrm{h}_{K}(\alpha )} {[K:\mathbb{Q}]} = \frac{\mathrm{h}_{M}(\alpha )} {[M:\mathbb{Q}]} = \frac{\mathrm{h}_{L}(\alpha )} {[L:\mathbb{Q}]}\,,}$$

which proves (B.2) in the general case. □ 

Now let α be an algebraic number. We define the absolute logarithmic height (or, simply, height) of α by

$$\displaystyle{ \mathrm{h}(\alpha ) = \frac{\mathrm{h}_{K}(\alpha )} {[K:\mathbb{Q}]}\,, }$$

(B.3)

where K is any number field containing α. By Proposition B.1, the right-hand side of (B.3) depends only on α and is independent of the particular choice of the field K.

We have defined a function $\mathrm{h}:\bar{ \mathbb{Q}} \rightarrow \mathbb{R}$. Its properties are summarized in the following proposition.

Proposition B.2.

• 1. For any $\alpha \in \bar{ \mathbb{Q}}$ we have h (α) ≥ 0.

• 2. ( Kronecker’s first theorem ) We have h (α) = 0 if and only if α = 0 or α is a root of unity.

• 3. ( Kronecker’s second theorem ) For every positive integer d, there exists $\varepsilon (d) > 0$ with the following property. Let α be an algebraic number of degree not exceeding d. Then either h (α) = 0 (in which case α is zero or a root of unity) or $\mathrm{h}(\alpha ) \geq \varepsilon (d)$ .

• 4. ( Galois action ) If α and β are conjugate over  $\mathbb{Q}$ then $\mathrm{h}(\alpha ) = \mathrm{h}(\beta )$ .

• 5. ( Height of a rational number ) Assume that $\alpha = a/b \in \mathbb{Q}$ , where $a,b \in \mathbb{Z}$ and gcd (a,b) = 1. Then

  $$\displaystyle{\mathrm{h}(\alpha ) =\log \max \{ \vert a\vert,\vert b\vert \}\,.}$$

  In particular, if α is a nonzero rational integer then $\mathrm{h}(\alpha ) =\log \vert \alpha \vert$ .

• 6. ( Height of a quotient ) More generally, let K be a number field, α,β ∈ K, and β ≠ 0. Then

  $$\displaystyle{ \mathrm{h}(\alpha /\beta ) = \frac{1} {[K:\mathbb{Q}]}\sum _{v\in M_{K}}\log \max \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\}. }$$

  (B.4)

  Also, if α and β are both algebraic integers, then

  $$\displaystyle{ \mathrm{h}(\alpha /\beta ) \leq \frac{1} {[K:\mathbb{Q}]}\sum _{\sigma:K\rightarrow \mathbb{C}}\log \max \{\vert \sigma (\alpha )\vert,\vert \sigma (\beta )\vert \}, }$$

  (B.5)

  the sum being extended to all embeddings $\sigma:K \rightarrow \mathbb{C}$ .

• 7. ( Heights of sums and products ) For any $\alpha,\beta \in \bar{ \mathbb{Q}}$ we have

  $$\displaystyle{ \mathrm{h}(\alpha \beta ) \leq \mathrm{h}(\alpha ) + \mathrm{h}(\beta ),\quad \mathrm{h}(\alpha +\beta ) \leq \mathrm{h}(\alpha ) + \mathrm{h}(\beta ) +\log 2. }$$

  (B.6)

  More generally, for $\alpha _{1},\ldots,\alpha _{n} \in \bar{ \mathbb{Q}}$ , we have

  $$\displaystyle\begin{array}{rcl} \mathrm{h}(\alpha _{1}\cdots \alpha _{n}) \leq \mathrm{h}(\alpha ) + \cdots + \mathrm{h}(\alpha _{n}),& & {}\end{array}$$

  (B.7)
  $$\displaystyle\begin{array}{rcl} \mathrm{h}(\alpha _{1} + \cdots +\alpha _{n}) \leq \mathrm{h}(\alpha _{1}) + \cdots + \mathrm{h}(\alpha _{n}) +\log n.& & {}\end{array}$$

  (B.8)

• 8. ( Height of a power ) For any $\alpha \in \bar{ \mathbb{Q}}$ and $n \in \mathbb{Z}$ (with α ≠ 0 if n < 0) we have

  $$\displaystyle{ \mathrm{h}(\alpha ^{n}) = \vert n\vert \mathrm{h}(\alpha )\,. }$$

  (B.9)

• 9. ( Height of a linear fraction ) Let $\left [\begin{array}{*{10}c} a&b\\ c &d\end{array} \right ]$ be a nondegenerate matrix with algebraic entries. Then for any algebraic number x ≠ − d∕c

  $$\displaystyle{\mathrm{h}\left (\frac{ax + b} {cx + d}\right ) = \mathrm{h}(x) + O(1),}$$

  where the constant implied by O(1) may depend on a,b,c,d (but not on x).

• 10. ( Northcott’s finiteness theorem ) For any C > 0 there exist only finitely many algebraic numbers α of degree and height bounded by C.

Proof.

Part 1 is obvious. To prove Part 4, put $K = \mathbb{Q}(\alpha )$, $L = \mathbb{Q}(\beta )$ and let τ: K → L be the isomorphism defined by τ(α) = β. As we have seen in Sect. A.8, the isomorphism τ induces a bijection $\tau _{{\ast}}: M_{K} \rightarrow M_{L}$, and, by the definition of τ _∗, we have $\vert \alpha \vert _{v} = \vert \beta \vert _{\tau _{{\ast}}v}$ for any v ∈ M _K . It follows that

$$\displaystyle{\mathrm{h}_{L}(\beta ) =\sum _{v\in M_{L}}\log \|\beta \|_{v} =\sum _{v\in M_{K}}\log \|\beta \|_{\tau _{{\ast}}v} =\sum _{v\in M_{K}}\log \|\alpha \|_{v} = \mathrm{h}_{K}(\alpha )\,.}$$

Since $[K:\mathbb{Q}] = [L:\mathbb{Q}]$, this implies h(α) = h(β).

For Part 6, observe that $\|\alpha /\beta \|_{v} = \vert \beta \vert _{v}^{-1}\max \left \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\right \}$. Hence

$$\displaystyle{\mathrm{h}_{K}(\alpha /\beta ) =\sum _{v\in M_{K}}\log \max \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\} -\log \prod _{v\in M_{K}}\vert \beta \vert _{v},}$$

and the second term vanishes by the product formula. This proves (B.4).

If α and β are algebraic integers, then $\log \max \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\} \leq 0$ for any finite v. Hence, (B.4) implies that

$$\displaystyle{ \mathrm{h}(\alpha /\beta ) \leq \frac{1} {[K:\mathbb{Q}]}\sum _{{v\in M_{K}\above 0.0pt v\mid \infty } }\log \max \{\vert \alpha \vert _{v},\vert \beta \vert _{v}\}\,, }$$

(B.10)

where the sum extends to infinite valuations of K. It remains to notice that the right-hand side of (B.10) is equal to the right-hand side of (B.5).

Part 5 [which is a reformulation of identity (B.1)] is a particular case of (B.4).

To prove Part 7, fix a number field K containing $\alpha _{1},\ldots,\alpha _{n}$ and observe that for any v ∈ M _K

$$\displaystyle\begin{array}{rcl} & \|\alpha _{1}\cdots \alpha _{n}\|_{v} \leq \|\alpha _{1}\|_{v}\cdots \|\alpha _{n}\|_{v}\,, & {}\\ & \|\alpha _{1} + \cdots +\alpha _{n}\|_{v} \leq \| n\|_{v}\max \left \{\|\alpha _{1}\|_{v},\ldots,\|\alpha _{n}\|_{v}\right \} \leq \| n\|_{v}\|\alpha _{1}\|_{v}\cdots \|\alpha _{n}\|_{v}\,.& {}\\ \end{array}$$

Taking the logarithm and summing up over v ∈ M _K , we obtain

$$\displaystyle\begin{array}{rcl} \mathrm{h}_{K}(\alpha _{1}\cdots \alpha _{n})& \leq & \mathrm{h}_{K}(\alpha ) + \cdots + \mathrm{h}_{K}(\alpha _{n}), {}\\ \mathrm{h}_{K}(\alpha _{1} + \cdots +\alpha _{n})& \leq & \mathrm{h}_{K}(\alpha _{1}) + \cdots + \mathrm{h}_{K}(\alpha _{n}) + \mathrm{h}_{K}(n). {}\\ \end{array}$$

It remains to divide by $[K:\mathbb{Q}]$ and to observe that h(n) = logn.

Part 8 for n > 0 is obvious, because $\|\alpha ^{n}\|_{v} =\|\alpha \|_{ v}^{n}$. To extend (B.9) to negative n, it suffices to observe that h(1∕α) = h(α) by (B.4).

In Part 9 put

$$\displaystyle{y = \frac{ax + b} {cx + d}.}$$

If c = 0 then the statement is immediate from Part 7. If c ≠ 0 then

$$\displaystyle{y = \frac{a} {c} + \frac{bc - ad} {cx + d} \cdot \frac{1} {c},}$$

and, using Parts 7 and 8, we find h(y) ≤ h(x) + O(1). Writing

$$\displaystyle{x = \frac{-dy + b} {cy - a},}$$

we prove similarly that h(x) ≤ h(y) + O(1), and Part 9 follows.

To prove Part 10, fix an algebraic number α of degree and height bounded by C. Let $x^{n} + a_{n-1}x^{n-1} + \cdots + a_{0}$ be the minimal polynomial of α over $\mathbb{Q}$. Using Parts 4 and 7, we may estimate the heights of the coefficients $a_{0},\ldots,a_{n-1}$ in terms of C. By Part 5, the numerators and denominators of the rational numbers $a_{0},\ldots,a_{n-1}$ are bounded in terms of C. Hence we have only finitely many possibilities for $a_{0},\ldots,a_{n-1}$. This proves Northcott’s theorem.

We are left with Parts 2 and 3 (Kronecker’s theorems). Obviously, the height of 0 is 0, as well as the height of any root of unity. Conversely, let α be an algebraic number of height 0. Then its powers 1, α, α ², … are of height 0 as well. Northcott’s theorem implies that among the numbers $1,\alpha,\alpha ^{2},\ldots,$ only finitely many are distinct. Hence there exist distinct integers ℓ and m such that $\alpha ^{\ell} =\alpha ^{m}$. This implies that α = 0 or α is a root of unity, proving Kronecker’s first theorem.

Kronecker’s second theorem is proved similarly. Let N = N(d) be the number of algebraic numbers of degree bounded by d and height bounded by 1. If h(α) ≠ 0 then all the powers 1, α, α ², … are pairwise distinct, which implies that h(α ^N) > 1. Hence Kronecker’s second theorem holds with $\varepsilon (d) = N(d)^{-1}$. □ 

The following statement is almost trivial, but it plays an important role in Diophantine analysis and, in particular, in the solution of Catalan’s problem.

Proposition B.3 (Liouville’s inequality).

Let K be a number field and α ∈ K ^× . Then for any v ∈ M _K we have

$$\displaystyle{ \vert \alpha \vert _{v} \geq \mathrm{e}^{-[K:\mathbb{Q}]\mathrm{h}(\alpha )}. }$$

(B.11)

More generally, for any $S \subseteq M_{K}$ , we have

$$\displaystyle{ \prod _{v\in S}\vert \alpha \vert _{v} \geq \mathrm{e}^{-[K:\mathbb{Q}]\mathrm{h}(\alpha )}. }$$

(B.12)

In particular, assume that K is a subfield of  $\mathbb{C}$ . Then

$$\displaystyle{ \vert \alpha \vert ^{d} \geq \mathrm{e}^{-[K:\mathbb{Q}]\mathrm{h}(\alpha )}, }$$

(B.13)

where d = 1 if $K \subset \mathbb{R}$ and d = 2 otherwise.

Proof.

We have

$$\displaystyle\begin{array}{rcl} \left (\prod _{v\in S}\vert \alpha \vert _{v}\right )^{-1} =\prod _{ v\in S}\left \vert \alpha ^{-1}\right \vert _{ v} \leq \prod _{v\in S}\left \|\alpha ^{-1}\right \|_{ v} \leq \prod _{v\in M_{K}}\left \|\alpha ^{-1}\right \|_{ v} = \mathrm{e}^{[K:\mathbb{Q}]\mathrm{h}(\alpha ^{-1}) }.& & {}\\ \end{array}$$

Since $\mathrm{h}(\alpha ^{-1}) = \mathrm{h}(\alpha )$, this proves (B.12) and, a fortiori, (B.11). For (B.13), let v ∈ M _K be the infinite valuation corresponding to the identical embedding $K\hookrightarrow \mathbb{C}$. Then | α |  _v  =  | α | if the embedding is real (that is, $K \subset \mathbb{R}$) and $\vert \alpha \vert _{v} = \vert \alpha \vert ^{2}$ otherwise. Applying (B.11) with this v, we obtain (B.13). □ 

Remark B.4.

Liouville’s inequality goes back to the celebrated work of Liouville [74] (published the same year as Catalan’s note extraite). Liouville proved that algebraic numbers cannot be well approximated by rationals. Precisely, if α is a real algebraic number of degree n > 1 and p∕q is a rational number, then

$$\displaystyle{ \left \vert \alpha -p/q\right \vert \geq c\vert q\vert ^{-n}, }$$

(B.14)

where c is a positive constant depending on α. One may deduce (B.14) from Proposition B.3 by applying (B.13) with α − p∕q instead of α and estimating h(α − p∕q) using Proposition B.2.

Appendix C: Commutative Rings, Modules, and Semi-simplicity

In this appendix, we recall several basic facts about rings and modules over them. In particular, we give a very brief treatise of semi-simplicity. The notion of semi-simplicity is central in the theory of noncommutative rings. In this outline we assume that the rings in question are commutative. This is irrelevant in some cases: for instance, the contents of Sect. C.3 extend without changes to the (left) modules over noncommutative base rings. On the other hand, the structure theory of semi-simple commutative rings (see Theorem C.11) is drastically simpler than its noncommutative counterpart.

All rings in this appendix are commutative and with 1. We assume that the reader is familiar with the definitions and basic facts about modules over a (commutative) ring, which can be found in any reasonable textbook of algebra.

One piece of notation: if S is a subset of an R-module, then the annihilator of S is, by definition, the set of all a ∈ R such that aS = 0. It is an ideal of R, which will be denoted by ann _R (S) or simply by ann(S). An R-module M is called exact if ann(M) = 0. Every R-module M admits a natural structure of an exact R∕ann(M)-module and, more generally, of $R/\mathfrak{a}$-module, where $\mathfrak{a}$ is any ideal annihilating M. If M′ is another R-module annihilated by $\mathfrak{a}$ then M is R-isomorphic to M′ if and only if M is $R/\mathfrak{a}$-isomorphic to M′.

Warning. In this appendix the abelian group law on modules is written additively, and the ring action is written multiplicatively. This is different from the rest of the book, where we normally use the multiplicative notation for the abelian group law and the exponential notation for the ring action.

C.1 Cyclic Modules

Let R be a (commutative) ring. An R-module M is called cyclic if it is generated over R by a single element. That is, there exists g ∈ M such that M = Rg.

If M = Rg is a cyclic module, then ann(M) = ann(g), and M is isomorphic, as an R-module, to the quotient ring R∕ann(M). In particular, the annihilator of a cyclic module defines it up to an isomorphism.

Since the submodules of R∕ann(M) are the ideals of the ring R∕ann(M), we have a one-to-one correspondence between the submodules of M and the ideals of R∕ann(M), or the ideals of R containing ann(M). Explicitly, the latter correspondence can be described as follows: if $\mathfrak{a}$ is an ideal of R containing ann(M) then the corresponding submodule of M is $\mathfrak{a}g$; conversely, if N is submodule of M then the corresponding ideal $\mathfrak{a}$ consists of all a ∈ R such that ag ∈ N.

If N is a submodule of a cyclic module M = Rg then the quotient M∕N is a cyclic module generated by the image of g. Also, if R is a principal ideal ring then so is R∕ann(M). We obtain the following statement.

Proposition C.1.

A quotient of a cyclic R-module is cyclic as well. If R is a principal ideal ring, then every submodule of a cyclic R-module is cyclic.

Maximal ideals of R∕ann(M) correspond to maximal (proper) submodules of M. Since every ring has a maximal ideal, we obtain the following assertion, which will be used in Sect. C.3.

Proposition C.2.

Every cyclic module has a proper maximal submodule.

This statement does not extend to arbitrary modules; for instance, the additive group $\mathbb{Q}$, viewed as a $\mathbb{Z}$-module, does not have a proper maximal submodule.

C.2 Finitely Generated Modules

Let R be a commutative ring. If M is a cyclic R-module and $\mathfrak{a}$ an ideal of R, then the following property is immediate: a ∈ R satisfies $aM \subset \mathfrak{a}M$ if and only if $a \in \mathfrak{a} + \mathrm{ann}(M)$. We want to extend this to finitely generated modules.

Theorem C.3.

Let M be a finitely generated R-module and let  $\mathfrak{a}$ be an ideal of R. Further, let a ∈ R satisfy $aM \subseteq \mathfrak{a}M$ . Then $a^{m} \in \mathfrak{a} + \mathrm{ann}(M)$ for some positive integer m.

(It will follow from the proof that m does not exceed the minimal number of generators of M. In particular, for the cyclic modules, we recover the statement above.)

For the proof, we need a lemma.

Lemma C.4.

Let M be an R-module generated by its elements $g_{1},\ldots,g_{m}$ and let $A = [\alpha _{ij}]_{1\leq i,j\leq m}$ be an m × m-matrix over R. Assume that

$$\displaystyle{ \sum _{j=1}^{m}\alpha _{ ij}g_{j} = 0\qquad (i = 1,\ldots,m). }$$

(C.1)

Then $\det A \in \mathrm{ann}(M)$ .

Proof.

Rewrite (C.1) as AG = 0, where G is the column $[g_{1},\ldots,g_{m}]$, and multiply this from the left by the adjoint matrix⁸ of A. We obtain the equality (detA)G = 0, that is, $(\det A)g_{1} = \cdots = (\det A)g_{m} = 0$. Hence detA annihilates M. □ 

Proof of Theorem C.3.

Let $g_{1},\ldots,g_{m}$ be a system of R-generators of M. Since every element of M is an R-linear combination of $g_{1},\ldots,g_{m}$, every element of $\mathfrak{a}M$ is an $\mathfrak{a}$-linear combination of $g_{1},\ldots,g_{m}$. In particular, if $aM \subseteq \mathfrak{a}M$ then there exists a matrix $A = [\alpha _{ij}]_{1\leq i,j\leq m}$ with entries in $\mathfrak{a}$ such that

$$\displaystyle{ag_{i} =\sum _{ j=1}^{m}\alpha _{ ij}g_{j}\qquad (i = 1,\ldots,m).}$$

Applying Lemma C.4, we find that $\det (aI - A) \in \mathrm{ann}(M)$. On the other hand, since the entries of A belong to $\mathfrak{a}$, we have $\det (aI - A) \equiv a^{m}\ \mathrm{mod}\,\mathfrak{a}$. This implies that $a^{m} \in \mathfrak{a} + \mathrm{ann}(M)$. □ 

If M is an R-module, and $\mathfrak{a}$ is an ideal of R, then $\bar{M} = M/\mathfrak{a}M$ has the natural structure of a module over the quotient ring $\bar{R} = R/\mathfrak{a}$. Obviously, if a ∈ R annihilates M, then its image in $\bar{R}$ annihilates $\bar{M}$. It is natural to ask whether the converse is true, that is, whether every $\bar{a} \in \bar{ R}$, annihilating $\bar{M}$, is an image of some a ∈ R annihilating M. In other words, we ask whether $\mathrm{ann}_{\bar{R}}(\bar{M})$ is the $\bar{R}$-image of ann _R (M). Of course, this is not true in general, but Theorem C.3 provides a sufficient condition for this.

Recall that $\mathfrak{a}$ is a radical ideal of R if for any a ∈ R and for any positive integer m we have $a^{m} \in \mathfrak{a} \Rightarrow a \in \mathfrak{a}$. Equivalently, $\mathfrak{a}$ is a radical ideal if the quotient ring $R/\mathfrak{a}$ has no nilpotent elements.

Corollary C.5.

Let M be a finitely generated R-module, and let  $\mathfrak{a}$ be an ideal of R. Put, as above, $\bar{R} = R/\mathfrak{a}$ and $\bar{M} = M/\mathfrak{a}M$ . Assume that $\mathfrak{a} + \mathrm{ann}_{R}(M)$ is a radical ideal of R. Then $\mathrm{ann}_{\bar{R}}(\bar{M})$ is the $\bar{R}$ -image of ann_R (M).

Proof.

As we have seen above, the image of ann _R (M) is contained in $\mathrm{ann}_{\bar{R}}(\bar{M})$. Conversely, let $\bar{a} \in \bar{ R}$ annihilate $\bar{M}$ , and let a ∈ R be a pullback of $\bar{\alpha }$. Then $aM \subset \mathfrak{a}M$, and Theorem C.3 implies that $a^{m} \in \mathfrak{a} + \mathrm{ann}_{R}(M)$ for some positive integer m. Since $\mathfrak{a} + \mathrm{ann}_{R}(M)$ is a radical ideal, we obtain $a \in \mathfrak{a} + \mathrm{ann}_{R}(M)$. Write a = α +β with $\alpha \in \mathfrak{a}$ and $\beta \in \mathrm{ann}_{R}(M)$. Then the image of β in $\bar{R}$ is $\bar{a}$. □ 

C.3 Semi-simple Modules

In this section we closely follow Lang [58, Sect. 17.2].

Let R be a (commutative) ring. A nonzero R-module is called simple if it has no submodules except 0 and itself.

For example, if R is a field then the simple modules are exactly the vector spaces of dimension 1. Also, simple $\mathbb{Z}$-modules are cyclic groups of prime order.

Simple modules are rather “rigid” objects. For instance, a morphism of a simple module (into another module) is either injective or a zero map. (Indeed, its kernel is a submodule of our simple module; hence it is either 0 or the module itself.) This property is usually called “Schur’s lemma.”

Another evidence for this “rigidness” is the fact that a sum of simple modules can always be made direct.⁹ More precisely, we have the following property.

Proposition C.6.

Let $\sum _{\lambda \in \varLambda }M_{\lambda }$ be a sum of simple submodules of a certain R-module. Then there is a subset $\varLambda ' \subseteq \varLambda$ such that $\sum _{\lambda \in \varLambda }M_{\lambda } =\bigoplus _{\lambda '\in \varLambda '}M_{\lambda '}$ .

Proof.

Let Λ′ be a maximal subset of Λ such that the sum $\sum _{\lambda '\in \varLambda '}M_{\lambda '}$ is direct. We have to show that every M _λ is contained in $M:=\bigoplus _{\lambda '\in \varLambda '}M_{\lambda '}$. We cannot have $M_{\lambda } \cap M = 0$ because otherwise we can add λ to the set Λ′ and still have a direct sum, contradicting the maximal choice of Λ′. Thus, $M_{\lambda } \cap M\neq 0$, and, since M _λ is simple, we have $M_{\lambda } \cap M = M_{\lambda }$, that is, $M_{\lambda } \subset M$. □ 

Now we are ready to define semi-simple modules.

Proposition C.7.

Let R be a commutative ring and M a module over R. Then the following three properties are equivalent:

• 1. The module M is a sum of its simple submodules.

• 2. The module M is a direct sum of its simple submodules.

• 3. Each submodule of M has a direct complement. (That is, if N is a submodule of M then there exists another submodule N′ such that M = N ⊕ N′.)

An R-module with these properties is called semi-simple.

Proof.

Implication (1) ⇒ (2) is Proposition C.6. To deduce implication (2) ⇒ (3), write $M =\bigoplus _{\lambda \in \varLambda }M_{\lambda }$, where each M _λ is simple, and let N be a submodule of M. Let Λ′ be a maximal subset of Λ such that the sum $N +\bigoplus _{\lambda '\in \varLambda '}M_{\lambda '}$ is direct. Arguing as in the proof of Proposition C.6, we show that $M = N \oplus \bigoplus _{\lambda '\in \varLambda '}M_{\lambda '}$.

We are left with the implication (3) ⇒ (1). First of all, let us show that Property 3 implies the following: every nonzero submodule of M has a simple submodule.

Thus, let N be a nonzero submodule of M. Since every nonzero module contains a cyclic submodule, we may assume that N is cyclic. By Proposition C.2, it has a maximal submodule S. By Property 3, the module S has a direct complement in M; that is, M = S ⊕ S′, where S′ is yet another submodule of M. It follows that N = S ⊕ P, where P = N ∩ S′. The module P, which is isomorphic to the quotient N∕S, is simple, because S is a maximal submodule of N.

Now we are ready to deduce the implication (3) ⇒ (1). Let M′ be the sum of all simple submodules of M. If M′ ≠ M then, by Property 3, we have M = M′ ⊕ M″, where M″ is a nonzero submodule. But, as we have just seen, M″ has a simple submodule, which contradicts our definition of M′ as the sum of all simple submodules. Thus, M′ = M. □ 

Remark C.8.

Property 3 of semi-simple modules can also be restated as follows: if M is a semi-simple module and N is a submodule of M then $M\mathop{\cong}N \oplus M/N$.

Semi-simplicity is inherited by sub- and quotient modules.

Proposition C.9.

Submodules and quotient modules of a semi-simple module are semi-simple.

Proof.

Let M be a semi-simple module and N its submodule. Further, let S be a submodule of N. Since M is semi-simple, S has a direct complement in M; that is, M = S ⊕ S′. It follows that $N = S \oplus (S' \cap N)$. We have shown that every submodule of N has a direct complement in N. Hence N is semi-simple.

Further, write M = N ⊕ N′. Then $M/N\mathop{\cong}N'$. Since N′ is semi-simple (as a submodule of M), so is M∕N. □ 

C.4 Semi-simple Rings

A ring is called semi-simple if it is semi-simple as a module over itself. Since a submodule of R is an ideal of R, and a simple submodule of R is a minimal (nonzero) ideal, the following conditions are equivalent by Proposition C.7:

• R is semi-simple;

• R is a sum of its minimal ideals;

• R is a direct sum of its minimal ideals;

• Every ideal $\mathfrak{a}$ of R has a direct complement (that is, there is an ideal $\mathfrak{a}'$ such that $R = \mathfrak{a} \oplus \mathfrak{a}'$).

For instance, a field is a semi-simple ring. Further, a direct product of finitely many semi-simple rings is again a semi-simple ring.

Indeed, it suffices to verify that a direct product of two semi-simple rings is semi-simple. Thus, let R ₁ and R ₂ be semi-simple rings, and let $\mathfrak{a}$ be an ideal of $R = R_{1} \times R_{2}$. Denote by $\mathfrak{a}_{1}$ and $\mathfrak{a}_{2}$ the projections of $\mathfrak{a}$ on R ₁ and R ₂, respectively, and let $\mathfrak{a}_{1}'$, respectively, $\mathfrak{a}_{2}'$ be a direct complement of $\mathfrak{a}_{1}$ in $R_{1}$, and, respectively, of $\mathfrak{a}_{2}$ in R ₂. Then a straightforward verification shows that $R = \mathfrak{a} \oplus \mathfrak{a}'$, where $\mathfrak{a}' = \mathfrak{a}_{1}' \times \mathfrak{a}_{2}'$. Hence R is semi-simple.

In particular, a direct product of finitely many fields is a semi-simple ring.

On the other hand, $\mathbb{Z}$ is not a semi-simple ring: the submodule $2\mathbb{Z}$ has no direct complement.

Proposition C.10.

Any module over a semi-simple ring is semi-simple.

Proof.

If R is a semi-simple ring then any free R-module is semi-simple, because it is a direct sum of modules isomorphic to R. Since any R-module is a quotient of a free R-module, it is semi-simple by Proposition C.9. □ 

It is remarkable that semi-simple commutative rings admit a very explicit classification.

Theorem C.11.

A (commutative) ring is semi-simple if and only if it is isomorphic to a direct product of finitely many fields.

Proof.

The “if” part is already proved in the beginning of the section. Now assume that R is a semi-simple ring and prove that it is isomorphic to a direct product of fields.

Write $R =\bigoplus _{\lambda \in \varLambda }\mathfrak{a}_{\lambda }$, where every $\mathfrak{a}_{\lambda }$ is a minimal nonzero ideal of R. It follows that every $x \in R$ can be (uniquely) presented as $x =\sum _{\lambda \in \varLambda }x_{\lambda }$, where $x_{\lambda } \in \mathfrak{a}_{\lambda }$, and for all but finitely many λ we have x _λ  = 0.

In particular, write $1 =\sum _{\lambda \in \varLambda }1_{\lambda }$ and let Λ′ be the finite subset of Λ consisting of λ-s such that 1 _λ ≠ 0. Then $1 =\sum _{\lambda \in \varLambda '}1_{\lambda }$, and, in particular, 1 belongs to the ideal $\bigoplus _{\lambda \in \varLambda '}\mathfrak{a}_{\lambda }$. However, 1 cannot belong to a proper ideal. Hence $\bigoplus _{\lambda \in \varLambda '}\mathfrak{a}_{\lambda } = R$, that is, Λ′ = Λ. Thus, we have proved that the set Λ is finite.

Further, since the sum is direct, we have $\mathfrak{a}_{\lambda } \cap \mathfrak{a}_{\mu } = 0$ for any distinct λ, μ ∈ Λ. Hence $\mathfrak{a}_{\lambda }\mathfrak{a}_{\mu } = 0$ for λ ≠ μ. In particular, if $x \in \mathfrak{a}_{\lambda }$ then 1 _μ x = 0 for any μ ≠ λ. It follows that for $x \in \mathfrak{a}_{\lambda }$,

$$\displaystyle{1_{\lambda }x = 1_{\lambda }x +\sum _{{\mu \in \varLambda \above 0.0pt \mu \neq \lambda } }1_{\mu }x = 1x = x.}$$

Thus, every $\mathfrak{a}_{\lambda }$ is a ring, with 1 _λ serving as its unity, and R is isomorphic to the direct product of the rings $\mathfrak{a}_{\lambda }$.

Moreover, let $\mathfrak{b}$ be an ideal of the ring $\mathfrak{a}_{\lambda }$, so that $\mathfrak{a}_{\lambda }\mathfrak{b} \subseteq \mathfrak{b}$. Then

$$\displaystyle{R\mathfrak{b} = \mathfrak{a}_{\lambda }\mathfrak{b} +\sum _{{\mu \in \varLambda \above 0.0pt \mu \neq \lambda } }\mathfrak{a}_{\mu }\mathfrak{b} = \mathfrak{a}_{\lambda }\mathfrak{b} \subseteq \mathfrak{b};}$$

that is, $\mathfrak{b}$ is an ideal of R as well. Since $\mathfrak{a}_{\lambda }$ is a minimal ideal of R, we have $\mathfrak{b} = \mathfrak{a}_{\lambda }$ or $\mathfrak{b} = 0$.

Thus, $\mathfrak{a}_{\lambda }$ is a ring without nontrivial ideals. Hence $\mathfrak{a}_{\lambda }$ is a field. The proof is complete. □ 

It is easy to describe the ideals of a semi-simple ring. Let R be a semi-simple ring, and write it as a direct product of finitely many fields: $R = K_{1} \times \cdots \times K_{s}$. Put Λ = { 1, …, s}. For λ ∈ Λ we denote by 1 _λ the element $(x_{1},\ldots,x_{s})$ such that x _λ  = 1 and x _μ  = 0 for μ ≠ λ (so that

$$\displaystyle{ 1_{1} = (1,0,\ldots,0),\qquad 1_{2} = (0,1,0,\ldots,0), }$$

(C.2)

etc.).

We leave to the reader the proof of the following proposition.

Proposition C.12.

• 1. For Λ′ ⊂Λ let  $\mathfrak{a}_{\varLambda '}$ consist of $x = (x_{1},\ldots,x_{s}) \in R$ such that x _λ = 0 for all λ∉Λ′. Then  $\mathfrak{a}_{\varLambda '}$ is an ideal of R. Conversely, any ideal of R is equal to  $\mathfrak{a}_{\varLambda '}$ for some $\varLambda ' \subseteq \varLambda$ .

• 2. The ideal  $\mathfrak{a}_{\varLambda '}$ principal; it is generated by the element $1_{\varLambda '} =\sum _{\lambda \in \varLambda '}1_{\lambda }$ . More generally,  $\mathfrak{a}_{\varLambda '}$ is generated by any $x = (x_{1},\ldots,x_{s})$ such that x _λ ≠ 0 for all λ ∈Λ′ (and x _λ = 0 for all λ∉Λ′).

• 3. In addition to this,  $\mathfrak{a}_{\varLambda '}$ is itself a (semi-simple) ring, with the unity 1 _Λ′ .

• 4. If Λ′ and Λ″ are subsets of Λ then

  $$\displaystyle{\mathfrak{a}_{\varLambda '}\mathfrak{a}_{\varLambda ''} = \mathfrak{a}_{\varLambda '\cap \varLambda ''},\qquad \mathfrak{a}_{\varLambda '} + \mathfrak{a}_{\varLambda ''} = \mathfrak{a}_{\varLambda '\cup \varLambda ''}.}$$

Corollary C.13.

A semi-simple ring is a principal ideal ring without nilpotent elements. It has only finitely many ideals. Every prime ideal of a semi-simple ring is maximal. Any quotient of a semi-simple ring is again a semi-simple ring. Any ideals $\mathfrak{a},\mathfrak{b}$ of a semi-simple ring satisfy $\mathfrak{a}\mathfrak{b} = \mathfrak{a} \cap \mathfrak{b}$ .

Another consequence of Proposition C.12 is that the direct complement of an ideal is defined uniquely.

Proposition C.14.

• 1. Let R be a semi-simple ring and  $\mathfrak{a}$ an ideal of R. Then there is a uniquely defined ideal  $\mathfrak{a}^{\perp }$ such that $\mathfrak{a} \oplus \mathfrak{a}^{\perp } = R$ .

• 2. For any ideals $\mathfrak{a},\mathfrak{b}$ of R, one has

  $$\displaystyle{ \left (\mathfrak{a}\mathfrak{b}\right )^{\perp } = \mathfrak{a}^{\perp } + \mathfrak{b}^{\perp },\quad \left (\mathfrak{a} + \mathfrak{b}\right )^{\perp } = \mathfrak{a}^{\perp }\mathfrak{b}^{\perp }. }$$

  (C.3)

  Also, $\mathfrak{a}\mathfrak{b} = 0$ if and only if $\mathfrak{b} \subseteq \mathfrak{a}^{\perp }$ .

The ideal $\mathfrak{a}^{\perp }$ will be called the complementary ideal, or, simply, the complement of $\mathfrak{a}$.

Proof.

Write $\mathfrak{a}$ as $\mathfrak{a}_{\varLambda '}$ and define $\mathfrak{a}^{\perp }$ as $\mathfrak{a}_{\varLambda \setminus \varLambda '}$. Then all the statements of this proposition are immediate consequences of Proposition C.12(4). □ 

It is equally easy to characterize R-modules. Let $R = K_{1} \times \cdots \times K_{s}$ be a semi-simple ring, and fix vector spaces V ₁ over K ₁,…, V _s over K _s . Then the direct product $M = V _{1} \times \cdots \times V _{s}$ is an R-module, the R-action being defined componentwise: if $a = (a_{1},\ldots,a_{s}) \in R$ and $v = (v_{1},\ldots,v_{s}) \in M$, then $av = (a_{1}v_{1},\ldots,a_{s}v_{s})$. It is easy to show that any R-module is of this type: for free modules this is obvious, and an arbitrary module is quotient of a free module.

Below we state several simple properties of modules over semi-simple rings. The proofs are left to the reader.

Proposition C.15.

Let $R = K_{1} \times \cdots \times K_{s}$ be a semi-simple ring and let $M = V _{1} \times \cdots \times V _{s}$ be an R-module. We put Λ ={ 1,…,s}:

• 1. The module M is finitely generated if and only if all the vector spaces $V _{1},\ldots,V _{s}$ are finite dimensional.

• 2. The module M is cyclic if and only if $\dim _{K_{\lambda }}V _{\lambda } \leq 1$ for all λ ∈Λ.

• 3. The annihilator of M is  $\mathfrak{a}_{\varLambda '}$ , where the set Λ′ consists of all λ ∈Λ with $\dim _{K_{\lambda }}V _{\lambda } = 0$ .

We have the following consequence for finite semi-simple rings.

Corollary C.16.

Let R be a finite semi-simple ring and let M be a finitely generated R-module. Then |M|≥|R∕ann (M)|, with equality if and only if M is cyclic.

Proof.

Write $R = K_{1} \times \cdots \times K_{s}$ and $M = V _{1} \times \cdots \times V _{s}$, and let d _λ be the dimension of the K _λ -vector space V _λ . Then

$$\displaystyle{\vert M\vert =\sum _{\lambda \in \varLambda }d_{\lambda }\vert K_{\lambda }\vert,\qquad \vert R/\mathrm{ann}(M)\vert =\sum _{{ \lambda \in \varLambda \above 0.0pt d_{\lambda }>0} }\vert K_{\lambda }\vert.}$$

It is clear that the second sum does not exceed the first sum, and the two are equal only if d _λ  ≤ 1 for all λ. □ 

C.5 The “Dual” Module

Let R be a commutative ring, S its subring, M an R-module, and N an S-module. Denote by Hom _S (M, N) the set of S-morphisms M → N. Obviously, Hom _S (M, N) has a natural S-module structure. But one can say more: Hom _S (M, N) has an R-module structure, defined as follows: given f ∈ Hom _S (M, N) and a ∈ R, the morphism af ∈ Hom _S (M, N) is defined by (af)(x) = f(ax) for x ∈ M. (Precisely speaking, if M is a left R-module, then Hom _S (M, N) becomes a right R-module. However, since the ring R is commutative, we need not distinguish between left and right.)

Now assume that the ring R has a field K as a subring. Then every R-module M is a K-vector space, and the “dual space” $M^{{\ast}} = \mathrm{Hom}_{K}(M,K)$ has a natural R-module structure, obtained by applying the previous paragraph with S = N = K. If our ring R is of finite K-dimension, and M is a finitely generated R-module, then M and the “dual module” M ^∗ are finite-dimensional K-vector spaces of the same dimension; in particular, they are K-isomorphic. We shall see that when R is semi-simple, they are R-isomorphic as well.

Theorem C.17.

Let R be a semi-simple (commutative) ring, containing a field K as a subring. Assume that R is of finite dimension over K, and let M be a finitely generated R-module. Then $M^{{\ast}} = \mathrm{Hom}_{K}(M,K)$ is R-isomorphic to M.

Proof.

Write R as a direct product of finitely many fields: $R = L_{1} \times \cdots \times L_{s}$. Each field L _λ contains a subfield isomorphic to K, given by K1 _λ (see (C.2) for the definition of 1 _λ ). Since R is finite dimensional over K, each L _λ is a finite extension of K.

Now write $M = V _{1} \times \cdots \times V _{s}$, where each V _λ is a finite-dimensional L _λ -vector space. Each $V _{\lambda }^{{\ast}} = \mathrm{Hom}_{K}(V _{\lambda },K)$ has an L _λ -module structure and is L _λ -isomorphic to V _λ , as follows by counting dimensions:

$$\displaystyle{\dim _{L_{\lambda }}V _{\lambda }^{{\ast}} = \frac{\dim _{K}V _{\lambda }^{{\ast}}} {[L_{\lambda }: K]} = \frac{\dim _{K}V _{\lambda }} {[L_{\lambda }: K]} =\dim _{L_{\lambda }}V _{\lambda }.}$$

Hence $V _{1}^{{\ast}}\times \cdots \times V _{s}^{{\ast}}$ is R-isomorphic to $M = V _{1} \times \cdots \times V _{s}$. Finally, the map

$$\displaystyle\begin{array}{rcl} M^{{\ast}}& \rightarrow & V _{ 1}^{{\ast}}\times \cdots \times V _{ S}^{{\ast}} {}\\ f& \mapsto & \left (f\vert _{V _{1}},\ldots,f\vert _{V _{s}}\right ) {}\\ \end{array}$$

is an R-isomorphism. The theorem is proved. □ 

Appendix D: Group Rings and Characters

In this appendix we collect basic facts about group rings and characters of finite commutative groups.

Let A be a commutative ring and G a finite group. The group ring A[G] is, by definition, the set of formal linear combinations $\sum _{g\in G}a_{g}g$, where a _g  ∈ A, with the operations defined in the obvious way:

$$\displaystyle\begin{array}{rcl} & \sum _{g\in G}a_{g}g +\sum _{g\in G}b_{g}g =\sum _{g\in G}(a_{g} + b_{g})g\,, & {}\\ & \left (\sum _{g\in G}a_{g}g\right )\left (\sum _{g\in G}b_{g}g\right ) =\sum _{g\in G}\left (\sum _{{h,k\in G\above 0.0pt hk=g} }a_{h}b_{k}\right )g\,.& {}\\ \end{array}$$

The unity of the group ring A[G] is e, the neutral element of the group G. This suggests the following convention: in the group ring, we identify e and 1. Precisely speaking, we may write a typical element of the group ring A[G] in both ways $\sum _{g\in G}a_{g}g$ or $a_{e} +\sum _{{g\in G\above 0.0pt g\neq e} }a_{g}g$, and we choose the writing most fit for the circumstances. In particular, we assume that A is a subring of A[G].

In this book the group G is usually abelian, and A is either the ring of integers $\mathbb{Z}$ or a field (except Appendix E, where $A = \mathbb{Z}/p\,^{s}\mathbb{Z}$ occurs).

D.1 The Weight Function and the Norm Element

In this section we recall the most basic notions about the group rings. We define the weight function w: A[G] → A by

$$\displaystyle{\mathrm{w}\left (\sum _{g\in G}a_{g}g\right ) =\sum _{g\in G}a_{g}.}$$

One immediately verifies that the weight function is additive and multiplicative:

Proposition D.1.

For any x,y ∈ A[G]

$$\displaystyle{\mathrm{w}(x + y) = \mathrm{w}(x) + \mathrm{w}(x),\qquad \mathrm{w}(xy) = \mathrm{w}(x)\mathrm{w}(y).}$$

Thus, the weight function is a ring homomorphism. Its kernel, consisting of elements of weight 0, is called the augmentation ideal of the group ring A[G].

The norm element of A[G] is

$$\displaystyle{\mathcal{N} =\sum _{g\in G}g.}$$

It is obvious that $x\mathcal{N} = \mathcal{N}x = \mathcal{N}$ for any x ∈ G. Extending this by linearity, we obtain the following property.

Proposition D.2.

For any x ∈ A[G] we have $x\mathcal{N} = \mathcal{N}x = \mathrm{w}(x)\mathcal{N}$ . In particular, $A[G]\mathcal{N} = \mathcal{N}A[G] = A\mathcal{N}$ .

The ideal $A\mathcal{N}$ is called the norm ideal of the group ring A[G]. If the cardinality  | G | is an invertible element of A (which is, in particular, the case if A is a field of characteristic not dividing  | G | ) then, writing each x ∈ A[G] as $(x -\mathrm{w}(x)\vert G\vert ^{-1}\mathcal{N}) + \mathrm{w}(x)\vert G\vert ^{-1}\mathcal{N}$, we obtain the following.

Proposition D.3.

Assume that |G| is an invertible element of A. Then A[G] is the direct sum of its augmentation ideal and its norm ideal.

The terms norm element and norm ideal are not common, but they are most suited for the purposes of this book.

D.2 Characters of a Finite Abelian Group

Let G be a (finite or infinite, commutative or not) group, and let K be a field. Denote by $\bar{K}$ the algebraic closure of K. A K-character ¹⁰ of G is a group homomorphism $\chi: G \rightarrow \bar{ K}^{\times }$. The character χ is called trivial if χ(g) = 1 for any g ∈ G. Characters form a multiplicative group, with the trivial character as the unity.

If G is finite, then the values of a nontrivial character sum to 0.

Proposition D.4.

Let χ be a nontrivial character of a finite group G. Then $\sum _{x\in G}\chi (x) = 0$ .

Proof.

Since χ is nontrivial, there exists g ∈ G such that χ(g) ≠ 1. We obtain

$$\displaystyle{0 =\sum _{x\in G}\chi (x) -\sum _{x\in G}\chi (gx) = (1 -\chi (g))\sum _{x\in G}\chi (x).}$$

Since 1 −χ(g) ≠ 0 by the choice of g, the result follows. □ 

Denote by $\bar{K}^{G}$ the $\bar{K}$-vector space of $\bar{K}$-valued functions on G. It is well known that the characters of G form a linearly independent subset of $\bar{K}^{G}$. This statement is usually attributed to Artin (see [58, Sect. 7.4]).

Proposition D.5 (Artin).

The K-characters of G are linearly independent over  $\bar{K}$ . That is, given pairwise distinct K-characters $\chi _{1},\ldots,\chi _{m}$ , and $\alpha _{1},\ldots,\alpha _{m} \in \bar{ K}$ , not all zero, the linear combination $\alpha _{1}\chi _{1} + \cdots +\alpha _{m}\chi _{m}$ is not identically zero on G.

Proof.

Assume that some nonzero linear combination $\alpha _{1}\chi _{1} + \cdots +\alpha _{m}\chi _{m}$ identically vanishes on G:

$$\displaystyle{ \alpha _{1}\chi _{1}(x) + \cdots +\alpha _{m}\chi _{m}(x) = 0 }$$

(D.1)

for any x ∈ G. We may assume that m is minimal; in particular, α ₁ ≠ 0. Also, m ≥ 2, and, in particular, $\chi _{1}\neq \chi _{m}$, because the characters are pairwise distinct. Thus, there exists g ∈ G such that $\chi _{1}(g)\neq \chi _{m}(g)$. Rewriting (D.1) with gx instead of x, we obtain

$$\displaystyle{ \alpha _{1}\chi _{1}(g)\chi _{1}(x) + \cdots +\alpha _{m}\chi _{m}(g)\chi _{m}(x) = 0. }$$

(D.2)

Multiplying (D.1) by χ _m (g) and subtracting the resulting identity from (D.2), we obtain

$$\displaystyle{ \beta _{1}\chi _{1}(x) + \cdots +\beta _{m-1}\chi _{m-1}(x) = 0 }$$

(D.3)

for any x ∈ G. Here $\beta _{k} =\alpha _{k}(\chi _{k}(g) -\chi _{m}(g))$, and, in particular,

$$\displaystyle{\beta _{1} =\alpha _{1}(\chi _{1}(g) -\chi _{m}(g))\neq 0.}$$

Identity (D.3) contradicts the minimal choice of m. The proposition is proved. □ 

In the sequel, unless the contrary is stated explicitly, the group G will be abelian and finite, and we shall assume that the characteristic of K does not divide |G|. Under these assumptions, the group of characters is usually called the dual group of G and is denoted by $\hat{G}$. The structure of this group is well known.

Theorem D.6.

Let G be a finite abelian group and K a field of characteristic not dividing |G|. Then the group of characters  $\hat{G}$ is isomorphic to G. In particular, there are exactly |G| distinct characters.

Proof.

We use induction on  | G | . Assume first that $G =\langle g\rangle$ is a cyclic group of finite order m. Then the map $\chi \mapsto \chi (g)$ is an isomorphism of $\hat{G}$ and the group of mth roots of unity in $\bar{K}$. Since char K does not divide m, the latter group is again cyclic of order m. This shows that $\hat{G}\mathop{\cong}G$.

Now let G be not cyclic. Then it is a direct sum of two nontrivial subgroups: $G = G_{1} \oplus G_{2}$. Consider the homomorphism $\hat{G} \rightarrow \hat{G}_{1} \times \hat{G}_{2}$ defined by $\chi \mapsto \left (\chi \left \vert _{G_{1}}\right.,\chi \left \vert _{G_{2}}\right.\right )$, and the homomorphism $\hat{G}_{1} \times \hat{G}_{2} \rightarrow \hat{G}$, which to each pair $(\chi _{1},\chi _{2})$ associates the character $\chi \in \hat{G}$ defined by $\chi (x_{1}x_{2}) =\chi _{1}(x_{1})\chi _{2}(x_{2})$. A routine verification shows that the two homomorphisms are inverse one to the other. Hence $\hat{G}\mathop{\cong}\hat{G}_{1} \times \hat{G}_{2}$. Since, by induction, $\hat{G}_{1}\mathop{\cong}G_{1}$ and $\hat{G}_{2}\mathop{\cong}G_{2}$, we obtain $\hat{G}\mathop{\cong}G$. □ 

Theorem D.6 implies that the characters form a basis of the space $\bar{K}^{G}$ of $\bar{K}$-valued functions on G: indeed, the characters are linearly independent, and their number is equal to the dimension | G | of this space. We obtain the following statement.

Proposition D.7.

Let G be a finite abelian group and K a field of characteristic not dividing |G|. Then the K-characters of the group G form a $\bar{K}$ -basis of the vector space  $\bar{K}^{G}$ . In particular, the |G|×|G| matrix $[\chi (g)]_{{ \chi \in \hat{G}\above 0.0pt g\in G} }$ is nondegenerate.

In the case $K \subseteq \mathbb{C}$ one can say more: the characters form an orthogonal basis of the space $\bar{K}^{G}$, with respect to the natural inner product

$$\displaystyle{(f_{1},f_{2}) = \frac{1} {\vert G\vert }\sum _{x\in G}f_{1}(x)\overline{f_{2}(x)}.}$$

Proposition D.8.

Assume that $K \subseteq \mathbb{C}$ . Then for any two characters χ ₁ and χ ₂ we have $(\chi _{1},\chi _{2}) = 1$ if $\chi _{1} =\chi _{2}$ and $(\chi _{1},\chi _{2}) = 0$ if $\chi _{1}\neq \chi _{2}$ .

Proof.

Since the values of any character χ are roots of unity, we have $\bar{\chi }=\chi ^{-1}$. Hence $\chi _{1}\bar{\chi _{2}}$ is a trivial character if and only if $\chi _{1} =\chi _{2}$. Applying Proposition D.4 to the character $\chi _{1}\bar{\chi _{2}}$, we obtain the result. □ 

D.3 Conjugate Characters

For a character $\chi \in \hat{G}$ we denote by K _χ the extension of K generated by the values of χ. It is the dth cyclotomic extension of K, where d is the order of χ. In particular, K _χ is a Galois extension of K. If σ ∈ Gal(K _χ ∕K), then σ ∘χ is also a K-character of G.

Let us say that two characters $\chi,\chi ' \in \hat{G}$ are conjugate (over K) if $K_{\chi } = K_{\chi '}$ and there exists $\sigma \in \mathrm{Gal}(K_{\chi }/K)$ such that χ′ = σ ∘χ. The conjugacy relation is an equivalence on $\hat{G}$, the class of every $\chi \in \hat{G}$ containing exactly $[K_{\chi }:K]$ elements. Thus, if we pick a representative in every conjugacy class of characters, and denote by M the set of chosen representatives, then we obtain the equality

$$\displaystyle{ \sum _{\chi \in M}[K_{\chi }:K] = \vert G\vert. }$$

(D.4)

Remark D.9.

Let us mention several simple facts, which are relevant here, but are not used in the book:

• 1. If χ and χ′ are conjugate characters, then they are of the same order; moreover, χ′ = χ ^a, where a is an integer coprime with the order.

• 2. The converse of the last statement is, in general, not true: for instance, if K is algebraically closed then the only character conjugate to χ is χ itself.

• 3. However, the converse is true if $K = \mathbb{Q}$; that is, if χ is a $\mathbb{Q}$-character of order m then for every $a \in \mathbb{Z}$, coprime with m, the character χ ^a is conjugate to χ.

• 4. In particular, two $\mathbb{Q}$-characters of a finite cyclic group are conjugate if and only if they are of the same order.

The proofs are left to the reader.

D.4 Semi-simplicity of the Group Ring

In this section G is a finite abelian group and K is a field of characteristic not dividing  | G | . The K-characters of G extend by linearity to the group ring K[G]: given $\chi \in \hat{G}$, we define the map K[G] → K _χ by

$$\displaystyle{\sum _{g\in G}a_{g}g\mapsto \sum _{g\in G}a_{g}\chi (g).}$$

As one immediately verifies, this is a ring homomorphism. We denote it also by χ, and we call it a character of the group ring K[G]. The set of all characters of K[G] will again be denoted by $\hat{G}$.

Non-degeneracy of the matrix $[\chi (g)]_{{ \chi \in \hat{G}\above 0.0pt g\in G} }$ (Proposition D.7) implies that the common kernel of all characters is trivial.

Proposition D.10.

If x ∈ K[G] satisfies χ(x) = 0 for every character $\chi \in \hat{G}$ , then x = 0.

Using this property, we show that the group ring K[G] is semi-simple, as defined in Appendix C.4.

Theorem D.11 (the “abelian Maschke theorem”).

Let G be a finite abelian group and K a field of characteristic not dividing |G|. Choose a system M of representatives of conjugacy classes of characters. Then the ring homomorphism

$$\displaystyle\begin{array}{rcl} [G]& \rightarrow & \prod _{\chi \in M}K_{\chi } \\ x& \mapsto & (\chi (x))_{\chi \in M}.{}\end{array}$$

(D.5)

is an isomorphism. In particular, the ring K[G] is semi-simple.

Proof.

Let x be in the kernel of the map (D.5), that is, χ(x) = 0 for any character χ ∈ M. Since conjugate characters vanish simultaneously at x, we obtain χ(x) = 0 for any $\chi \in \hat{G}$. Proposition D.10 implies that x = 0. Hence (D.5) is a monomorphism and, by (D.4), the K-dimensions of both parts of (D.5) are equal. Therefore we have an isomorphism. □ 

Remark D.12.

In the case “G is a cyclic group of order m and $K = \mathbb{Q}$” Theorem D.11 implies the isomorphism $\mathbb{Q}[G]\mathop{\cong}\prod _{d\mid m}\mathbb{Q}(\zeta _{d})$. We do not use this in the present book.

As Proposition C.12 suggests, the ideals of K[G] can be characterized as common kernels of characters from the set M: for a subset S of M let $\mathcal{I}_{S}$ be the common kernel of the characters from the complement $M\setminus S$. Then $\mathcal{I}_{S}$ is an ideal of K[G], and any ideal of K[G] is equal to $\mathcal{I}_{S}$ for some $S \subseteq M$.

The ideal $\mathcal{I}_{S}$ is isomorphic, as a $K[G]$-module, to $\prod _{\chi \in S}K_{\chi }$. It follows that it is a K-vector space, and

$$\displaystyle{\dim _{K}\mathcal{I}_{S} =\sum _{\chi \in S}[K_{\chi }: K].}$$

If θ ∈ K[G] then the principal ideal (θ) is equal to $\mathcal{I}_{S}$, where S consists of characters χ ∈ M with χ(θ) ≠ 0. Hence

$$\displaystyle{ \dim _{K}(\theta ) =\sum _{\chi (\theta )\neq 0}[K_{\chi }: K]. }$$

(D.6)

Since conjugate characters vanish at θ simultaneously, and since for every χ there are exactly [K _χ : K] characters, conjugate to χ, the sum in (D.6) is equal to the number of $\chi \in \hat{G}$ with χ(θ) ≠ 0. We have proved the following statement.

Proposition D.13.

Let θ be an element of K[G]. Then the K-dimension of the principal ideal (θ) is equal to the number of characters $\chi \in \hat{G}$ nonvanishing at θ.

Finally, we state one more consequence of Theorem D.11, to be used in Appendix F.4. Let N be a module over the group ring K[G]. Then N is a K-vector space, and the “dual space” $N^{{\ast}} = \mathrm{Hom}_{K}(N,K)$ has a natural G-module structure (see Appendix C.5). The following result is a direct application of Theorems C.17 and D.11.

Corollary D.14.

Let G be a finite abelian group and K a field of characteristic not dividing |G|. Let N be a finitely generated K[G]-module. Then the “dual module” $N^{{\ast}} = \mathrm{Hom}_{K}(N,K)$ is K[G]-isomorphic to N.

D.5 Idempotents

We retain the setup of the previous section. It is useful to have an explicit basis of the ideal $\mathcal{I}_{S}$ as a $K$-vector space and an explicit generator of $\mathcal{I}_{S}$ as a principal ideal. In this section we produce both, under the additional assumption

$$\displaystyle{ \mbox{ $K$ contains the $\vert G\vert $th roots of unity,} }$$

(D.7)

which is sufficient for our purposes. The reader is invited to examine the general case.

Assumption (D.7) implies that K _χ  = K for all characters χ, or, equivalently, every character is conjugate only to itself. It follows that in Theorem D.11 we have $M = \hat{G}$, and the map (D.5) becomes the isomorphism

$$\displaystyle{\psi: K[G] \rightarrow K^{\hat{G}},}$$

where $K^{\hat{G}}$ is the ring of K-functions on $\hat{G}$.

For every character χ, we define the element $\varepsilon _{\chi } \in K[G]$ as follows:

$$\displaystyle{\varepsilon _{\chi } = \frac{1} {\vert G\vert }\sum _{g\in G}\chi (g)g^{-1}.}$$

These elements have many remarkable properties; here are some of them.

Proposition D.15.

• 1. For any characters χ and χ′ we have

  $$\displaystyle{\chi '(\varepsilon _{\chi }) = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &\mbox{ if $\chi =\chi '$},\\ 0\quad &\mbox{ if $\chi \neq \chi '$}.\end{array} \right.}$$

• 2. For any x ∈ K[G] we have $x\varepsilon _{\chi } =\chi (x)\varepsilon _{\chi }$ .

• 3. For any χ we have $\varepsilon _{\chi }^{2} =\varepsilon _{\chi }$ .

Proof.

Applying Proposition D.4 to the character χ(χ′)⁻¹, we find

$$\displaystyle{\chi '(\varepsilon _{\chi }) = \frac{1} {\vert G\vert }\sum _{g\in G}\chi (g)\chi '(g)^{-1} = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &\mbox{ if $\chi =\chi '$}, \\ 0\quad &\mbox{ if $\chi \neq \chi '$}. \end{array} \right.}$$

This proves Part 1. Part 2 is trivially true for any x ∈ G. By linearity, it extends to x ∈ K[G]. Part 3 is an immediate consequence of the previous parts. □ 

The element $\varepsilon _{\chi }$ is called the idempotent of χ. Notice that the idempotent of the trivial character is $\vert G\vert ^{-1}\mathcal{N}$.

Since $x\varepsilon _{\chi } \in K\varepsilon _{\chi }$ for any x ∈ K[G], the set $K\varepsilon _{\chi }$ is an ideal of $K[G]$. We have decomposed K[G] into a direct sum of one-dimensional ideals:

$$\displaystyle{K[G] =\bigoplus _{\chi \in \hat{G}}K\varepsilon _{\chi }\,.}$$

The ideal $\mathcal{I}_{S}$ decomposes as

$$\displaystyle{\mathcal{I}_{S} =\bigoplus _{\chi \in S}K\varepsilon _{\chi }\,,}$$

which, in particular, implies that $\{\varepsilon _{\chi }:\chi \in S\}$ is a K-basis of $\mathcal{I}_{S}$.

Also, it is easy to see that

$$\displaystyle{\mathcal{I}_{S} = K[G]\sum _{\chi \in S}\varepsilon _{\chi },}$$

which gives an explicit generator of the principal ideal $\mathcal{I}_{S}$. Since we do not use this result, its proof is left to the reader.

Appendix E: Reduction and Torsion of Finite G-Modules

In this appendix we study the reduction of G-modules modulo a prime number. Like in Appendix C (and unlike in the rest of the book) here we write the abelian group law on modules additively and the ring action multiplicatively.

Let M be a finite $\mathbb{Z}$-module (written additively), and let p be a prime number. It is easy to see that M∕pM is isomorphic to the p-torsion submodule M[p ] = { a ∈ M: p  a = 1}. In this appendix we show that this automorphism extends, under a mild assumption, to finite G-modules.

Theorem E.1.

Let G be a finite abelian group and let M be a finite G-module. Further, let p be a prime number not dividing |G|. Then M∕pM and M[p ] are G-isomorphic.

It is easy to see (see end of Sect. E.4) that M can be replaced by its p-Sylow submodule and the group ring $\mathbb{Z}[G]$ can be replaced by the finite ring $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$ with sufficiently large s. We introduce a very special class of rings (called here telescopic rings) and describe, in terms of these rings, the structure of the group ring $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$ and of the finitely generated $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$-modules. After this preparation Theorem E.1 becomes immediate.

When writing this appendix we profited from very useful discussions with Jean Fresnel. Many arguments below are due to him or based on his ideas.

E.1 Telescopic Rings

Call a (commutative) ring R telescopic if R has an ideal $\mathfrak{m}$ such that any other ideal is a power of $\mathfrak{m}$. A basic example of a telescopic ring is $\mathbb{Z}/p\,^{s}\mathbb{Z}$, where p is a prime number: in this ring every ideal is a power of the principal ideal (p).

Obviously, R is a local ring with the maximal ideal $\mathfrak{m}$. Also, since the zero ideal is a power of $\mathfrak{m}$, the latter is nilpotent: $\mathfrak{m}^{s} = 0$ for some s. If we choose the minimal s with this property (called the index of nilpotency of $\mathfrak{m}$) then $\mathfrak{m}^{k}\neq \mathfrak{m}^{k+1}$ for k < s: indeed, if $\mathfrak{m}^{k} = \mathfrak{m}^{k+1}$ then, multiplying by $\mathfrak{m}^{s-k-1}$, we obtain $\mathfrak{m}^{s-1} = \mathfrak{m}^{s} = 0$, contradicting the minimal choice of s. It follows that

$$\displaystyle{ R = \mathfrak{m}^{0} \supsetneq \mathfrak{m} \supsetneq \ldots \supsetneq \mathfrak{m}^{s} = 0 }$$

(E.1)

is the complete list of ideals of the telescopic ring R. Thus, we may regard telescopic rings as a “nilpotent analogue” of discrete valuation rings.

It is important that the maximal ideal of a telescopic ring is principal (and hence so are all its ideals). Indeed, if s = 1 then $\mathfrak{m} = 0$ (in which case R is a field), and if s > 1 then $\mathfrak{m} \supsetneq \mathfrak{m}^{2}$, and we may choose an element $\pi \in \mathfrak{m}$ which does not belong to $\mathfrak{m}^{2}$. Then $\mathfrak{m} \supseteq \pi R \supsetneq \mathfrak{m}^{2}$, which shows that $\mathfrak{m} =\pi R$.

Conversely, telescopic rings can be characterized as rings having a principal and nilpotent maximal ideal.

Proposition E.2.

A commutative ring with a principal and nilpotent maximal ideal is telescopic.

Proof.

Let $\mathfrak{m}$ be a principal and nilpotent maximal ideal of a ring R. Since $\mathfrak{m}$ is nilpotent, it is contained in any other maximal ideal of R, which means that $\mathfrak{m}$ is the only maximal ideal of R. Thus, R is a local ring; in particular, every element outside $\mathfrak{m}$ is invertible.

It remains to show that every proper ideal of R is a power of $\mathfrak{m}$. Thus, let $\mathfrak{a}$ be a nonzero proper ideal of R, and let k be the maximal integer with the property $\mathfrak{a} \subseteq \mathfrak{m}^{k}$ (since $\mathfrak{m}$ is nilpotent, the set of integers with this property is finite). We are going to show that $\mathfrak{a} = \mathfrak{m}^{k}$.

By the maximal choice of k we have $\mathfrak{a}\not\subseteq \mathfrak{m}^{k+1}$. Let α be an element of $\mathfrak{a}$ not contained in $\mathfrak{m}^{k+1}$. Recall that $\mathfrak{m}$ is a principal ideal, and let π be its generator. Since α belongs to $\mathfrak{m}^{k}$, but not to $\mathfrak{m}^{k+1}$, we have $\alpha =\pi ^{k}\varepsilon$, where $\varepsilon \notin \mathfrak{m}$. Since R is a local ring, the element $\varepsilon$ is invertible. Thus, $\pi ^{k} \in \mathfrak{a}$, which proves that $\mathfrak{a} = \mathfrak{m}^{k}$. □ 

Remark E.3.

It is not sufficient to assume only that the maximal ideal is nilpotent (without assuming it principal), as shows the following example, kindly communicated to us by Jean Fresnel. Let K be a field, and put $R = K[x,y]/(x^{2},y^{2},xy)$. Then $\mathfrak{m} = xR + yR$ is a maximal ideal of R satisfying $\mathfrak{m}^{2} = 0$, but R is not a telescopic ring, because the ideal xR is not a power of $\mathfrak{m}$.

E.2 Products of Telescopic Rings

Let R be a direct product of finitely many telescopic rings: $R = R_{1} \times \cdots \times R_{n}$, where every R _j is telescopic. Let $\mathfrak{m}_{j} =\pi _{j}R_{j}$ be the maximal ideal of R _j and $K_{j} = R_{j}/\mathfrak{m}_{j}$ be the residue field. Then $\pi = (\pi _{1},\ldots,\pi _{n})$ is a nilpotent element of R, and R∕π R is isomorphic to $K_{1} \times \cdots \times K_{n}$. In particular, R∕π R is a semi-simple ring (see Appendix C.4).

It turns out that this property characterizes products of telescopic rings.

Theorem E.4.

Let R be a (commutative) ring, having a nilpotent element π such that the quotient ring R∕πR is semi-simple. Then R is a direct product of finitely many telescopic rings.

As a consequence, we obtain the structure of the group ring $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$.

Corollary E.5.

Let R be a telescopic ring with residue field K, and let G be a finite abelian group. Assume that the characteristic of K does not divide G. Then the group ring R[G] is a direct product of finitely many telescopic rings.

In particular, if G is a finite abelian group and p is a prime number not dividing |G|, then for any positive integer s the ring $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$ is a direct product of telescopic rings.

Proof.

Let π generate the maximal ideal of the telescopic ring R. Then R[G]∕π R[G] is isomorphic to K[G], which is a semi-simple ring by Theorem D.11. Whence the result. □ 

For the proof of Theorem E.4 we need a lemma (a familiar reader will quickly recognize in it a version of “Hensel’s lemma”).

Lemma E.6.

Let π be a nilpotent element of a ring R and let $\alpha _{0},\beta _{0}$ be coprime elements of R satisfying $\alpha _{0}\beta _{0} \equiv 0\ \mathrm{mod}\,\pi$ . Then there exists coprime $\alpha,\beta \in R$ such that $\alpha \equiv \alpha _{0}\ \mathrm{mod}\,\pi$ ,   $\beta \equiv \beta _{0}\ \mathrm{mod}\,\pi$ , and αβ = 0.

Proof.

Write $\alpha _{0}\beta _{0} =\pi \lambda$. Since α ₀ and β ₀ are coprime, there exist u, v ∈ R such that $u\alpha _{0} + v\beta _{0} = 1$. Put $\alpha _{1} =\alpha _{0} - v\lambda \pi$ and $\beta _{1} =\beta _{0} - u\lambda \pi$. Then $\alpha _{1}\beta _{1} \equiv 0\ \mathrm{mod}\,\pi ^{2}$. Also, α ₁ and β ₁ are coprime. Indeed, $u\alpha _{1} + v\beta _{1} \equiv 1\ \mathrm{mod}\,\pi$, and an element congruent to 1 modulo π is invertible¹¹.

Iterating the process, we find, for every k, coprime elements $\alpha _{k},\beta _{k} \in R$ such that $\alpha _{k} \equiv \alpha _{0}\ \mathrm{mod}\,\pi$,  $\beta _{k} \equiv \beta _{0}\ \mathrm{mod}\,\pi$, and $\alpha _{k}\beta _{k} \equiv 0\ \mathrm{mod}\,\pi ^{2^{k} }$. Since π is nilpotent, we shall eventually obtain $\alpha _{k}\beta _{k} = 0$. □ 

We shall also use the “Chinese remainder theorem”: if $\mathfrak{a}$ and $\mathfrak{b}$ are coprime ideals of a ring R, then the natural homomorphism $R \rightarrow R/\mathfrak{a} \times R/\mathfrak{b}$ is surjective and defines an isomorphism $R/\mathfrak{a}\mathfrak{b}\mathop{\cong}R/\mathfrak{a} \times R/\mathfrak{b}$. See, for instance, [58, Sect. 2.2] or [3, Proposition 1.10].¹²

Proof of Theorem E.4.

We write $R/\pi R = K_{1} \times \cdots \times K_{n}$, where $K_{1},\ldots,K_{n}$ are fields and, arguing by induction on n, we shall prove that R is a product of n telescopic rings. The case n = 1 is exactly Proposition E.2.

Now assume that n > 1. There exist coprime α, β ∈ R such that α β = 0, the image of α in $K_{1} \times \cdots \times K_{n}$ is e ₁ = (1, 0, …, 0) and the image of β is $1 - e_{1} = (0,1,\ldots,1)$. To find such α and β, fix an arbitrary α ₀ ∈ R with image e ₁, put $\beta _{0} = 1 -\alpha _{0}$, and apply Lemma E.6.

By the Chinese remainder theorem, the ring R is isomorphic to R′ × R″, where R′ = R∕α R and R″ = R∕β R. Let π′, respectively π″, be the image of π in R′, respectively R″. Then

$$\displaystyle{R'/\pi 'R' = R/(\alpha R +\pi R) = K_{2} \times \cdots \times K_{n},}$$

and by induction we conclude that R′ is a direct product of n − 1 telescopic rings. Similarly, R″∕π″R″ = K ₁, whence R″ is a telescopic ring. The theorem is proved. □ 

E.3 Elementary Divisors and Finitely Generated Modules

Like principal ideal domains, telescopic rings admit the “theory of elementary divisors,” and a finitely generated module over a telescopic ring is a direct sum of its cyclic submodules.

Theorem E.7.

Let n ≥ 1 be an integer, R a telescopic ring with the maximal ideal  $\mathfrak{m}$ , and H a submodule of the free module R ⁿ . Then there exist $a_{1},\ldots,a_{n} \in R^{n}$ such that

$$\displaystyle{ R^{n} = Ra_{ 1} \oplus \cdots \oplus Ra_{n},\qquad H = \mathfrak{m}^{r_{1} }a_{1} \oplus \cdots \oplus \mathfrak{m}^{r_{n} }a_{n}, }$$

(E.2)

where $r_{1},\ldots,r_{n}$ are nonnegative integers.

Proof.

Let $p_{k}: R^{n} \rightarrow R$ be the projection on the kth coordinate. Then $p_{k}(H) = \mathfrak{m}^{t_{k}}$, where $t_{1},\ldots,t_{k}$ are nonnegative integers. We put

$$\displaystyle{r_{1} =\min \{ t_{1},\ldots,t_{n}\}}$$

and we may assume that $r_{1} = t_{1}$, so that $p_{1}(H) = \mathfrak{m}^{r_{1}}$ and $p_{k}(H) \subseteq \mathfrak{m}^{r_{1}}$ for $k = 2,\ldots,n$.

Let π be a generator of $\mathfrak{m}$. Then H has an element b such that $p_{1}(b) =\pi ^{r_{1}}$. Write $b = \left (\pi ^{r_{1}},\pi ^{r_{1}}x_{2},\ldots,\pi ^{r_{1}}x_{n}\right ) =\pi ^{r_{1}}a_{1}$, where $a_{1} = (1,x_{1},\ldots,x_{n})$. Obviously, $R^{n} = Ra_{1} \oplus \ker p_{1}$, and an easy verification shows that $H = Rb \oplus H'$, where $H' =\ker p_{1} \cap H$. Since $Rb = \mathfrak{m}^{r_{1}}a_{1}$, we obtain

$$\displaystyle{R^{n} = Ra_{ 1} \oplus \ker p_{1},\qquad H = \mathfrak{m}^{r_{1} }a_{1} \oplus H'.}$$

By induction, there exist $a_{2},\ldots,a_{n} \in \ker p_{1}$ such that

$$\displaystyle{\ker p_{1} = Ra_{2} \oplus \cdots \oplus Ra_{n},\qquad H' = \mathfrak{m}^{r_{2} }a_{2} \oplus \cdots \oplus \mathfrak{m}^{r_{n} }a_{n}.}$$

This proves the theorem. □ 

Corollary E.8.

A finitely generated module over a telescopic ring is a direct sum of its cyclic submodules.

Proof.

Let M be a finitely generated module over a telescopic ring R. Then for some n there is a surjective homomorphism $R^{n} \rightarrow M$. Let H be its kernel, so that $M\mathop{\cong}R^{n}/H$, and let $a_{1},\ldots,a_{n} \in R^{n}$ be such that (E.2) holds. Then $M = R\alpha _{1} \oplus \cdots \oplus R\alpha _{n}$, where $\alpha _{1},\ldots,\alpha _{n} \in M$ are the images of $a_{1},\ldots,a_{n}$, respectively. □ 

E.4 Reduction and Torsion

Let R be a ring, let M be an R-module, and let $\mathfrak{a}$ be an ideal of R. Put $M[\mathfrak{a}] =\{ x \in M: \mathfrak{a}x = 0\}$, and call it the $\mathfrak{a}$ -torsion submodule of M.

Proposition E.9.

Let M be a finitely generated module over a telescopic ring R, and let  $\mathfrak{a}$ be an ideal of R. Then $M/\mathfrak{a}M$ is R-isomorphic to $M[\mathfrak{a}]$ .

Proof.

If $M =\bigoplus _{ i=1}^{n}M_{i}$ is a direct sum of its submodules, then

$$\displaystyle{M[\mathfrak{a}]\mathop{\cong}\bigoplus _{i=1}^{n}M_{ i}[\mathfrak{a}],\qquad M/\mathfrak{a}M\mathop{\cong}\bigoplus _{i=1}^{n}M_{ i}/\mathfrak{a}M_{i}.}$$

By Corollary E.8 this reduces Proposition E.9 to the case when M is a cyclic module. Replacing R by the ring $R/\mathrm{ann}_{R}(M)$ (and $\mathfrak{a}$ by its image in this ring), we may assume that $\mathrm{ann}_{R}(M) = 0$ and thereby M≅R.

Let $\mathfrak{m} =\pi R$ be the maximal ideal of R, and write $\mathfrak{a} = \mathfrak{m}^{k} =\pi ^{k}R$. Then $M/\mathfrak{a}M = R/\pi ^{k}R$ and $M[\mathfrak{a}] =\pi ^{s-k}R$, where s is the nilpotency index of π. The map R → R defined by $x\mapsto \pi ^{s-k}x$ has π ^s−k R as its image and π ^k R as its kernel. Hence $R/\pi ^{k}R\mathop{\cong}\pi ^{s-k}R$, as wanted. □ 

Proposition E.9 extends to direct products of telescopic rings.

Proposition E.10.

Let R be a direct product of finitely many telescopic rings, and let M be a finitely generated R-module. Then for any ideal  $\mathfrak{a}$ of  $R$ we have $M/\mathfrak{a}M\mathop{\cong}M[\mathfrak{a}]$ .

Proof.

Write $R = R_{1} \times \cdots \times R_{n}$, where each R _k is a telescopic ring. Put

$$\displaystyle{e_{1} = (1,0,\ldots,0),\qquad e_{2} = (0,1,0,\ldots,0),}$$

and so on, so that $1 = e_{1} + \cdots + e_{n}$. Then an R-module M splits into a direct sum as $M =\bigoplus _{ i=1}^{n}M_{i}$, where $M_{i} = e_{i}M$. As in the proof of Proposition E.9, it suffices to verify the required property for every M _i .

Since M _i is annihilated by 1 − e _i it can be viewed as a module over the ring R∕(1 − e _i )R, which is isomorphic to the telescopic ring R _i . We reduced Propositions E.10–E.9, hereby completing the proof. □ 

Now we are ready to prove Theorem E.1.

Proof of Theorem E.1.

Let M _p be the p-Sylow subgroup of M. Then M _p is a G-submodule of M, and the embedding $M_{p}\hookrightarrow M$ induces $G$-isomorphisms $M[p\,]\mathop{\cong}M_{p}[p\,]$ and $M/pM\mathop{\cong}M_{p}/pM_{p}$. Hence we may assume that M is a p-group; in particular, M has a natural structure of a $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$-module for a suitable s.

By the assumption, p does not divide  | G | . Hence $\mathbb{Z}/p\,^{s}\mathbb{Z}[G]$ is a direct product of finitely many telescopic rings (Corollary E.5), and our theorem becomes a direct consequence of Proposition E.10. □ 

Appendix F: Radical Extensions

In this appendix we give an account of the theory of the q-radical extensions, that is, the extensions generated by qth roots of elements of the field. Everywhere in this appendix K is a field and q a prime number, distinct from the characteristic of K.

F.1 Field Generated by a Single Root

Fix β ∈ K and a qth root β ^1∕q. Then all conjugates of β ^1∕q over K are contained in the set

$$\displaystyle{ \left \{\beta ^{1/q}\xi:\xi \in \upmu _{ q}\right \}, }$$

(F.1)

where $\upmu _{q} = \left \{1,\zeta _{q},\ldots,\zeta _{q}^{q-1}\right \}$ is the group of the qth roots of unity.

Proposition F.1.

If β is not a qth power in K, then [K(β ^1∕q ): K] = q for any choice of the root β ^1∕q .

In other words, the polynomial x ^q −β either has a root in K or is irreducible over K (in which case (F.1) is the full set of conjugates of β ^1∕q).

Proof.

Assume that [K(β ^1∕q): K] = r < q for some choice of the root β ^1∕q. We have $\mathcal{N}_{K(\beta ^{1/q})/K}(\beta ^{1/q}) =\beta ^{r/q}\xi$ with ξ ∈ μ _q . Since q is prime, there exist integers a and b such that ar + bq = 1. Then $\left ((\beta ^{r/q}\xi )^{a}\beta ^{b}\right )^{q} =\beta$; that is, β is a qth power in K. □ 

Since [K(ζ _q ): K] ≤ q − 1, we obtain the following consequence, which will be systematically used in the sequel.

Corollary F.2.

If β ∈ K is not a qth power in K, then it is not a qth power in K(ζ _q ) either.

F.2 Kummer’s Theory

We wish to study the Galois group (over K) of the field $K(\zeta _{q},\beta ^{1/q})$ (the field, generated by all the qth roots of β). Actually, we consider a more general situation. We fix a subgroup B of the multiplicative group K ^×, and consider the field $L = K(\root{q}\of{B})$, generated over K by the set

$$\displaystyle{\root{q}\of{B} = \left \{\rho \in \bar{ K}^{\times }:\rho ^{q} \in B\right \}.}$$

It is a Galois extension of K (the composite of splitting fields of all polynomials x ^q −β, where β ∈ B). The structure of the Galois group Γ = Gal(L∕K) depends on whether or not ζ _q belongs to K.

In this section we assume that

$$\displaystyle{ \zeta _{q} \in K. }$$

(F.2)

In this case the theory of q-radical extensions is called Kummer’s theory. Fix β ∈ B and a qth root β ^1∕q. Then all conjugates of β ^1∕q over K are contained in the set $\left \{\beta ^{1/q}\xi:\xi \in \upmu _{q}\right \}$, where

$$\displaystyle{\upmu _{q} = \left \{1,\zeta _{q},\ldots,\zeta _{q}^{q-1}\right \}}$$

is the group of the qth roots of unity. In other words, for any γ ∈ Γ, we have $(\beta ^{1/q})^{\gamma }/\beta ^{1/q} \in \upmu _{q}$. The quotient $(\beta ^{1/q})^{\gamma }/\beta ^{1/q}$ depends only on β and γ, but not on the particular choice of the qth root β ^1∕q; indeed, if we replace β ^1∕q by β ^1∕q ξ with ξ ∈ μ _q , the quotient is not changed because $\upmu _{q} \subset K$ by the assumption (F.2).

To continue, recall the notion of pairing. Given groups V and W, and an abelian group A, an A-pairing of V and W is a map f: V × W → A such that for any y ∈ W the map $x\mapsto f(x,y)$ is a group homomorphism V → A, and for any x ∈ V the map $y\mapsto f(x,y)$ is a group homomorphism W → A. The pairing induces group homomorphisms V → Hom(W, A) and W → Hom(V, A). The kernels of these homomorphisms are called left and right kernels of our pairing. The pairing is left (respectively, right) faithful if its left (right) kernel is 1 and faithful if it is both left and right faithful.

In our case we have Kummer’s pairing $B\times \varGamma \rightarrow \upmu _{q}$ defined by

$$\displaystyle{(\beta,\gamma )\mapsto \frac{{\bigl (\beta ^{1/q}\bigr )}^{\gamma }} {\beta ^{1/q}}.}$$

It is obviously right faithful, which implies that Γ is isomorphic to a subgroup of Hom(B, μ _q ). It follows that Γ is a q-torsion abelian group; in particular, it has a natural structure of an $\mathbb{F}_{q}$-vector space.

However, Kummer’s pairing is not left faithful in general, its right kernel being $B \cap (K^{\times })^{q}$. Putting $\bar{B} = B/B \cap (K^{\times })^{q}$, we obtain a faithful pairing $\bar{B}\times \varGamma \rightarrow \upmu _{q}$, and the following holds.

Theorem F.3.

If the group B is finitely generated, then Γ is isomorphic to   $\bar{B}$ .

This is a special case of a general statement in linear algebra. Let F be a field and V and W vector spaces over F. A pairing V × W → F is F-bilinear if both the induced maps V → Hom(W, F) and W → Hom(V, F) are F-linear.

Proposition F.4.

Assume that one of the spaces V and W is of finite F-dimension and that there exists a faithful F-bilinear pairing V × W → F; then V and W are isomorphic vector spaces.

Proof.

Assume, for instance, that V is finite dimensional. Then the dual space V ^∗ = Hom(V, F) is of the same dimension as V. Since our pairing is right faithful, we have a linear monomorphism $W\hookrightarrow V ^{{\ast}}$. Hence, W is finite dimensional, and

$$\displaystyle{\dim W \leq \dim V ^{{\ast}} =\dim V.}$$

Similarly, dimV ≤ dimW. Hence dimV = dimW, as wanted. □ 

Proof of Theorem F.3.

Both groups $\bar{B}$ and Γ have a natural structure of an $\mathbb{F}_{q}$-vector space. Identifying μ _q with the additive group of $\mathbb{F}_{q}$, we obtain a faithful $\mathbb{F}_{q}$-bilinear pairing $\bar{B}\times \varGamma \rightarrow \mathbb{F}_{q}$. Since B is finitely generated, $\bar{B}$ is a finite-dimensional $\mathbb{F}_{q}$-space. Now apply Proposition F.4. □ 

If B′ is a subgroup of $B(K^{\times })^{q}$ then the field $L' = K(\root{q}\of{B'})$ is a subfield of L. In the case when B is finitely generated, any subfield of L, containing K, is of this form. More precisely, we have the following statement (which will not be used in the sequel).

Proposition F.5.

Assume that B is finitely generated. Then there is a one-to-one correspondence between the group towers $(K^{\times })^{q} \leq B' \leq B(K^{\times })^{q}$ and the field towers $K \subseteq L' \subseteq L$ , given by $L' = K(\root{q}\of{B'})$ .

Proof.

If there is a faithful F-bilinear pairing $V \times W\stackrel{f}{\rightarrow }F$ of finite-dimensional vector spaces, then the subspaces $V '$ of V and W′ of W stay in the one-to-one correspondence given by¹³ W′ = (V ′)^⊥. In our case we obtain a one-to-one correspondence between the subgroups of $\bar{B}$ and of Γ. Since the former correspond to the group towers $(K^{\times })^{q} \leq B' \leq B(K^{\times })^{q}$ and the latter to the field towers $K \subseteq L' \subseteq L$, we obtain a one-to-one correspondence between the two types of towers. A straightforward inspection shows that $L' = K(\root{q}\of{B'})$ for the corresponding towers. □ 

Without assuming that B is finitely generated, the following statement still holds: there is a one-to-one correspondence between the group towers $(K^{\times })^{q} \leq B' \leq B(K^{\times })^{q}$ and the field towers $K \subseteq L' \subseteq L$ such that both the index $[B(K^{\times })^{q}: B']$ and the degree [L: L′] are finite, and the correspondence is again given by $L' = K(\root{q}\of{B'})$. We leave the proof to the reader.

F.3 General Radical Extensions

We no longer assume that ζ _q  ∈ K. Since 1 ∈ B, we have $\zeta _{q} \in \root{q}\of{B}$, which means that we have a tower of fields

$$\displaystyle{K \subset K(\zeta _{q}) \subset L,}$$

where $\zeta _{q}$ is a primitive qth root of unity and $L = K(\root{q}\of{B})$.

Proposition F.6.

Assume that the group B is finitely generated. Then the group $\varGamma = \mathrm{Gal}(L/K(\zeta _{q}))$ is isomorphic to $\bar{B} = B/B \cap (K^{\times })^{q}$ .

Proof.

Corollary F.2 implies that $B \cap (K^{\times })^{q} = B \cap (K(\zeta _{q})^{\times })^{q}$. Now apply Theorem F.3 with K(ζ _q ) instead of K. □ 

Putting Δ = Gal(L∕K) and H = Gal(K(ζ _q )∕K), we obtain the exact sequence

$$\displaystyle{ 1 \rightarrow \varGamma \rightarrow \varDelta \rightarrow H \rightarrow 1. }$$

(F.3)

Both extensions K(ζ _q )∕K and L∕K(ζ _q ) are abelian, but, as we shall see in a while, when ζ _q ∉K, the extension L∕K is badly non-abelian.

Theorem F.7.

Assume that ζ _q ∉K. Let L ₀ be an abelian extension of K contained in L. Then $L_{0} \subset K(\zeta _{q})$.

The proof relies on a simple group-theoretic lemma. To state it, we need some preparation. Recall, first of all, the notion of group extension. Let

$$\displaystyle{ 1 \rightarrow A \rightarrow E \rightarrow Q \rightarrow 1 }$$

(F.4)

be an exact sequence of groups, the group A being abelian. The group E acts on its normal subgroup A by conjugation (we write this action exponentially, $a\mapsto a^{e} = e^{-1}ae$). Since A is abelian, it acts on itself trivially. This induces a natural right action of Q = E∕A on A (again written exponentially).

Now assume that we are given an abelian group A, a group Q, and a right Q-action on A. Then any exact sequence (F.4) inducing the given Q-action on A is called an extension of Q by A (with respect to the given action).

Now let F be a field and A an F-vector space. It will be convenient for us to write the group structure on A multiplicatively and the F-action on A exponentially (that is, instead of the familiar λ x +μ y, we write $x^{\lambda }y^{\mu }$).

Lemma F.8.

Let Q be a subgroup of the multiplicative group F ^× , and let 1 → A → E → Q → 1 be an extension of Q by A with respect to the standard action of Q on A. Assume that Q ≠ 1. Then ¹⁴ A = [E,E], the commutator subgroup of E.

In other words, if E′ is a normal subgroup of E with abelian quotient E∕E′, then E′ ≥ A.

The assumption Q ≠ 1 is essential: if Q = 1 then E = A is an abelian group and [E, E] = 1.

Proof.

Since the quotient E∕A is abelian, we have [E, E] ≤ A, so it suffices to prove that A ≤ [E, E].

Recall that we write the group law on A multiplicatively and the F-action on A exponentially. Fix an element x ≠ 1 of Q. Then, when a runs over A, the expression a ^x−1 runs over A as well. If for every x ∈ Q we fix a lifting $\tilde{x} \in E$, then $a^{x} = \tilde{x}^{-1}a\tilde{x}$. It follows that

$$\displaystyle{a^{x-1} = a^{-1+x} = a^{-1}\tilde{x}^{-1}a\tilde{x} ={\bigl [ a,\tilde{x}\,\bigr ]}.}$$

Thus, every element of A can be presented as a commutator ${\bigl [a,\tilde{x}\,\bigr ]}$. Hence A ≤ [E, E]. □ 

Proof of Theorem F.7.

Consider the exact sequence (F.3). The group Γ is an $\mathbb{F}_{q}$-vector space (see Sect. F.2), and H can be viewed as a subgroup of $(\mathbb{Z}/q\mathbb{Z})^{\times } = \mathbb{F}_{q}^{\times }$, the action of H on Γ being exactly the (restricted to H) $\mathbb{F}_{q}$-vector space action. Since ζ _q  ⊄ K, we have H ≠ 1 and Lemma F.8 applies. By the lemma, every subgroup of Δ with abelian quotient contains Γ. By the Galois theory, this means that every subfield of L, abelian over K, is contained in the subfield fixed by Γ, that is, in K(ζ _q ). The theorem is proved. □ 

F.4 Equivariant Kummer’s Theory

Now assume that K is a Galois extension of some field K ₀, with Galois group $G = \mathrm{Gal}(K/K_{0})$, and that B is a G-invariant subgroup of K. Then $L = K(\root{q}\of{B})$ is a Galois extension of K ₀ (the composite of splitting fields of polynomials $\prod _{\sigma \in G}(x -\beta ^{\sigma })$, where β ∈ B). It follows that G acts naturally on the group Δ = Gal(L∕K), and the subgroup Γ = Gal(L∕K(ζ _q )) is invariant under this action.¹⁵

Thus, both B and Γ are $G$-modules, and so is $\bar{B} = B/B \cap (K^{\times })^{q}$. Recall that Γ and $\bar{B}$ are isomorphic as abelian groups when B is finitely generated. It is natural to ask whether they are isomorphic as G-modules. In general, this is not true. For instance, assume that $\zeta _{q}\notin K_{0}$, $K = K_{0}(\zeta _{q})$ and B ⊂ K ₀. Then B is a trivial G-module, but Γ is a nontrivial G-module.

Similar examples show that to have G-isomorphism we must assume that

$$\displaystyle{ K \cap K_{0}(\zeta _{q}) = K_{0}. }$$

(F.5)

Adding to these some technical hypotheses, we indeed prove that $\bar{B}$ and Γ are G-isomorphic.

Theorem F.9.

In the above setup, assume that the group B is finitely generated and that  (F.5) holds. Assume, in addition, that the group G is finite and abelian and that q does not divide |G|. Then  $\bar{B}$ and Γ are G-isomorphic.

First of all, we establish an equivariant version of Proposition F.4. Let V and W be G-modules, where G is a group, and let A be an abelian group. A pairing $V \times W\stackrel{f}{\rightarrow }A$ is called G-equivariant if for any x ∈ V, y ∈ W, and σ ∈ G, we have $f(x^{\sigma },y^{\sigma }) = f(x,y)$.

Proposition F.10.

Let F be a field, G a finite abelian group, V, W two finitely generated F[G]-modules, and V × W → F a G-equivariant faithful bilinear pairing. Assume that the characteristic of F does not divide |G|. Then V and W are isomorphic as F[G]-modules.

Proof.

According to Proposition F.4, the natural map W → V ^∗ is an isomorphism of vector spaces. Moreover, G-equivariance of the pairing implies that this map is a G-morphism. Thus, W is F[G]-isomorphic to the dual space V ^∗. Finally, Corollary D.14 implies that V ^∗ is G-isomorphic to V. □ 

Proof of Theorem F.9.

Assume first that $\zeta _{q} \in K_{0}$. Then we also have ζ _q  ∈ K, and Kummer’s theory (Sect. F.2) applies. In particular, we have Kummer’s pairing $B \times \varGamma \stackrel{f}{\rightarrow }\upmu _{q}$, and it is easy to verify that Kummer’s pairing is G-equivariant.

Indeed, for γ ∈ Γ and σ ∈ G, we have $\gamma ^{\sigma } = \tilde{\sigma }^{-1}\gamma \tilde{\sigma }$, where $\tilde{\sigma }$ is a lifting of σ to $\mathrm{Gal}(L/K_{0})$. Further, if β ^1∕q is a qth root of β ∈ B, then $\left (\beta ^{1/q}\right )^{\tilde{\sigma }}$ is a qth root of β ^σ. It follows that

$$\displaystyle{f(\beta ^{\sigma },\gamma ^{\sigma }) = \frac{\left (\beta ^{1/q}\right )^{\tilde{\sigma }\gamma ^{\sigma }}} {\left (\beta ^{1/q}\right )^{\tilde{\sigma }}} = \frac{\left (\beta ^{1/q}\right )^{\gamma \tilde{\sigma }}} {\left (\beta ^{1/q}\right )^{\tilde{\sigma }}} = f(\beta,\gamma )^{\tilde{\sigma }}.}$$

But $f(\beta,\gamma ) \in \upmu _{q} \subset K_{0}$ (recall that $\zeta _{q} \in K_{0}$), whence $f(\beta,\gamma )^{\tilde{\sigma }} = f(\beta,\gamma )$. Thus, we have proved that $f(\beta ^{\sigma },\gamma ^{\sigma }) = f(\beta,\gamma )$ for any β ∈ B, γ ∈ Γ, and σ ∈ G. Hence Kummer’s pairing is G-equivariant.

It follows that the faithful pairing $\bar{B}\times \varGamma \rightarrow \mathbb{F}_{q}$, defined in the proof of Theorem F.3, is G-equivariant as well. Proposition F.10 implies that $\bar{B}$ is F _q [G]-isomorphic to Γ. This proves the theorem in the special case $\zeta _{q} \in K_{0}$.

In the general case, assumption (F.5) implies that

$$\displaystyle{\mathrm{Gal}\left (K(\zeta _{q})/K_{0}(\zeta _{q})\right ) = G.}$$

Hence we may replace K ₀ by $K_{0}(\zeta _{q})$ and K by K(ζ _q ), reducing the general case to the case $\zeta _{q} \in K_{0}$, already proved. □ 

Remark F.11.

The results of this section remain true without assuming the group G abelian. The proofs are, basically, the same, but one should use the full (non-abelian) version of the Maschke theorem (Theorem D.11).