# Algebraic Numbers

## Abstract

Chapter 11 introduces algebraic number theory which developed through the attempts of mathematicians to prove Fermat’s Last Theorem. An algebraic number is a complex number which is algebraic over the field Q of rational numbers. An algebraic number field is a subfield of the field C of complex numbers, which is a finite field extension of the field Q and obtained from Q by adjoining a finite number of algebraic elements. The concepts of algebraic numbers, algebraic integers, Gaussian integers, algebraic number fields and quadratic fields are introduced in this chapter after a short discussion on general properties of field extension and finite fields. There are several proofs of Fundamental Theorem of Algebra. It is proved in this chapter by using homotopy (discussed in Chap. ). Moreover, countability of algebraic numbers, existence of transcendental numbers, impossibility of duplication of a general cube and that of trisection of a general angle are shown in this chapter.

## Keywords

Finite Field Fundamental Theorem Field Extension Algebraic Number Algebraic Integer
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

Algebraic number theory arose through the attempts of mathematicians to prove Fermat’s Last Theorem. One of the continuous themes since early 20th century which motivated algebraic number theory was to establish analogy between algebraic number fields and algebraic function fields. The study of algebraic number theory was initiated by many mathematicians. This list includes the names of Kronecker, Kummer, Dedekind, Dirichlet, Gauss, and many others. Gauss called algebraic number theory ‘Queen of Mathematics’. Andrew Wiles established Fermat’s Last Theorem a few years back. An algebraic number is a complex number which is algebraic over the field Q of rational numbers. An algebraic number field is a subfield of the field C of complex numbers, which is a finite field extension of the field Q and is obtained from Q by adjoining a finite number of algebraic elements. The concepts of algebraic numbers, algebraic integers, Gaussian integers, algebraic number fields and quadratic fields are introduced in this chapter after a short discussion on general properties of field extensions and finite fields. Moreover, the countability of algebraic numbers, existence of transcendental numbers, impossibility of duplication of general cube and impossibility of trisection of a general angle by straight edge and compass are shown. The celebrated theorem known as Fundamental Theorem of Algebra is also proved in this chapter by using the tools from homotopy theory as discussed in Chap. . This theorem proves the algebraic completeness of the the field of complex numbers. Like the field of complex numbers, we further prove that the field of algebraic numbers is also algebraically closed.

## 11.1 Field Extension

We begin with a review of basic concepts of field extension followed by finite fields. While studying field extension, we mainly consider a pair of fields F and K such that F is a subfield of K. Taking F as the basic field, an extension field K of F is a field K which contains F as a subfield. The basic results needed for the study of field extensions are discussed first followed by a discussion on simple extensions.

## Definition 11.1.1

Let K be a field and F be a subfield of K. Then K is said to be a field extension of F, written as K/F.

## Example 11.1.1

(i) C is a field extension of R, denoted by C/R.

(ii) $$\mathbf{Q}(\sqrt{2})=\{a+b\sqrt{2}: a,b\in \mathbf{Q}\}$$ is a field extension of Q, denoted by $$\mathbf{Q}(\sqrt{2})/\mathbf{Q}$$.

(iii) R is a field extension of Q, denoted by R/Q.

(iv) Let K be a field. If $$\operatorname{char} K$$ is 0, then K contains a subfield F isomorphic to Q. If $$\operatorname{char} K=p>0$$, for some prime p, then K contains a subfield F isomorphic to Z p . Thus K can be viewed as a field extension of F, where F is a field isomorphic to Q or Z p according as $$\operatorname{char} K$$ is 0 or p.

(v) Let F be a field and K=F(x), the field of rational functions in x over F, which is the quotient field of the integral domain F[x]. Then K is a field extension of F.

## Definition 11.1.2

Let K be a field extension of F and α 1,α 2,…,α n be in K. Then the smallest field extension of F containing both F and {α 1,α 2,…,α n } is called the field generated by F and {α 1,α 2,…,α n } and is denoted by F(α 1,α 2,…,α n ).

We now describe the elements of F(α 1,α 2,…,α n ).

## Theorem 11.1.1

Let K be a field extension of F and α 1,α 2,…,α n be in K. Then
\begin{aligned} &F(\alpha_1,\alpha_2,\ldots,\alpha_n) \\ &\quad= \bigl\{ f(\alpha_1,\alpha_2,\ldots, \alpha_n)/g(\alpha_1,\alpha_2,\ldots , \alpha_n): f,g\in F[x_1,x_2, \ldots,x_n]\textit{ and } \\ &\qquad\ g(\alpha_1,\alpha_2, \ldots,\alpha_n)\neq0\bigr\} . \end{aligned}

## Proof

Let S be the set defined by S={f(α 1,α 2,…,α n )/g(α 1,α 2,…,α n ):f,gF[x 1,x 2,…,x n ] and g(α 1,α 2,…,α n )≠0}. Then S is a subfield of K such that FS, since a=a/1∈S for every aF. Let L be a subfield of K containing F and {α 1,α 2,…,α n }. Then LS, because L contains f(α 1,α 2,…,α n ) and also f(α 1,α 2,…,α n )/g(α 1,α 2,…,α n ), if g(α 1,α 2,…,α n )≠0. This shows that S is the smallest subfield of K containing F and {α 1,α 2,…,α n }. Hence F(α 1,α 2,…,α n )=S. □

## Corollary

Let K be a field extension of F and αK. Then F(α)={f(α)/g(α), where f(x),g(x)∈F[x] and g(α)≠0} is the quotient field of F[α].

## Definition 11.1.3

Let K be a field extension of F. The field K is said to be a simple extension of F iff there exists an element α in K such that K=F(α). The element αK is said to be a primitive element of the extension and F(α) is said to be generated by F and α.

## Example 11.1.2

The field extension $$\mathbf{Q}(\sqrt{5})$$ of Q is a simple extension with $$\sqrt{5}$$ as its primitive element. The field $$\mathbf{Q}(\sqrt{5})$$ is generated by Q and a root $$\sqrt{5}$$ of the equation x 2−5=0 and consists of all real numbers $$a+b \sqrt{5}$$ with rational coefficients a and b.

## Definition 11.1.4

A field F is said to be a prime field iff F has no proper subfield.

## Example 11.1.3

(i) Q is a prime field, because Q has no proper subfield.

(ii) Z p is a prime field for every prime integer p.

## Definition 11.1.5

Let K be a field extension of the field F. Then an element α of K is said to be
1. (a)

a root of a polynomial f(x)=a 0+a 1 x+⋯+a n x n in F[x] iff f(α)=a 0+a 1 α+⋯+a n α n =0;

2. (b)

algebraic over F iff α is a root of some non-null polynomial f(x) in F[x];

3. (c)

transcendental over F iff α is not a root of any non-null polynomial in F[x].

## Remark

Every element a of a field F is algebraic over F.

## Definition 11.1.6

Let K be a field extension of the field F. Then K is said to be an algebraic extension over F iff every element of K is algebraic over F. Otherwise, K is called a transcendental extension over F. A simple extension F(α) is said to be an algebraic or transcendental over F according to whether the element α is algebraic or transcendental over F.

## Example 11.1.4

1. (a)
1. (i)

For the field extension $$\mathbf{Q}(\sqrt{2})$$ of Q, every element $$\alpha=a+b\sqrt{2}\in\mathbf{Q}(\sqrt{2})$$ is algebraic over Q.

2. (ii)

The element iC is algebraic over R.

3. (iii)

The element π in R is transcendental over $$\mathbf{Q}(\sqrt{2})$$.

4. (iv)

The element 2πiC is algebraic over R but transcendental over Q.

2. (b)
For the field extension R of Q,
1. (i)

e and π are both transcendental over Q;

For proof see [Hardy and Wright (2008, pp. 218–227)];

2. (ii)

$$\sqrt{\pi}$$ is transcendental over Q;

3. (iii)

π 2 is transcendental over Q.

## Remark

Example 11.1.4(a)(iv) shows that the two properties algebraic and transcendental vary depending on the base field.

## Theorem 11.1.2

Let F be a field and F(x) be the field of rational functions in x over F. Then the element x of F(x) is transcendental over F.

## Proof

F(x) is clearly a field extension of the field F. If x is not transcendental over F, then there is a non-null polynomial f(y)=a 0+a 1 y+⋯+a n y n over F, of which x is a root. This shows that a 0+a 1 x+⋯+a n x n =0. Hence each a i =0 implies that f(y) is a null polynomial. This gives a contradiction. □

## Theorem 11.1.3

Let K be a field extension of F and an element α of K be algebraic over F. Then F(α) and F[x]/〈f(x)〉 are isomorphic for some monic irreducible polynomial f(x) in F[x], of which α is a root.

## Proof

Define the mapping μ:F[x]→F[α],f(x)↦f(α).

Then μ is an epimorphism and hence by the First Isomorphism Theorem, F[x]/kerμF[α]. But α is algebraic over F⇒kerμ≠{0}. Again kerμ is an ideal of the Euclidean domain F[x]. But F[x] is a PID ⇒kerμ is a principal ideal of F[x]. Hence kerμ=〈g(x)〉 for some g(x)∈F[x]. If a is the leading coefficient of g(x), then f(x)=a −1 g(x) is a monic polynomial of F[x] and kerμ=〈g(x)〉=〈f(x)〉. Clearly, f(x) is irreducible in F[x] and hence 〈f(x)〉 is a maximal ideal in F[x]. Consequently, F[x]/〈f(x)〉 is a field. Thus F(α)=Q(F[α])≅Q(F[x]/〈f(x)〉)=F[x]/〈f(x)〉. Hence the theorem follows. □

## Corollary

Let K be a field extension of the field F. If an element α of K is algebraic over F, then there exists a unique monic irreducible polynomial in F[x] of which α is a root.

## Proof

Using Theorem 11.1.3, there exists a monic irreducible polynomial f(x) in F[x] such α is a root of f(x). If g(x) is a monic irreducible polynomial in F[x] such that α is a root of g(x), then by the above definition of μ,g(x)∈kerμ=〈f(x)〉. Consequently, f(x) divides g(x) in F[x]. Suppose g(x)=q(x)f(x) for some q(x)∈F[x]. Since g(x) is irreducible in F[x], it follows that either q(x) or f(x) is a unit in F[x]. As f(x) is not a unit in F[x], q(x) is a unit and thus q(x)=c for some non-zero element of F. Again, since f(x) and g(x) are both monic, c=1 and hence f(x)=g(x). □

This corollary leads to Definition 11.1.7.

## Definition 11.1.7

Let F be a field and K be a field extension of F. Then the unique monic irreducible polynomial f(x) in F[x] having α as a root is called the minimal polynomial of α over F and the degree of f(x) is called the degree of α over F.

We now show that, under a suitable condition, F(α)=F[α].

## Proposition 11.1.1

Let K be a field extension of the field F. Then F(α)=F[α] iff α is algebraic over F.

## Proof

Suppose α is algebraic over F. Then as proved in Theorem 11.1.3, F[α]≅F[x]/〈f(x)〉, where f(x) is a monic irreducible polynomial in F[x], of which α is a root. Since F[x]/〈f(x)〉 is a field, F[α] is a field and hence F(α)=F[α].

Conversely, suppose F(α)=F[α]. If α=0, then α is the root of the polynomial xF[x]. If α≠0, then α −1F(α) and hence α −1=a 0+a 1 α+⋯+a n α n for some a i F, i=0,1,…,n, nN. Consequently, 0=−1+a 0 α+a 1 α 2+⋯+a n α n+1α is a root of the non-null polynomial f(x)=−1+a 0 x+a 1 x 2+⋯+a n x n+1F[x]⇒α is algebraic over F. □

## Example 11.1.5

(i) The element i of C is algebraic over R and x 2+1 is the minimal polynomial of i over R. Again xi is the minimal polynomial of i over C.

(ii) The element $$\sqrt{5}$$ of $$\mathbf{Q}(\sqrt{5})$$ is algebraic over Q and x 2−5 is the minimal polynomial of $$\sqrt{5}$$ over Q. The degree of $$\sqrt{5}$$ over Q is 2.

## Theorem 11.1.4

Let K be a field extension of the field F and the element α of K is algebraic over F. Suppose f(x) is the minimal polynomial of α over F.
1. (a)

If a polynomial g(x) of F[x] has a root α, then f(x) divides g(x) in F[x].

2. (b)

f(x) is the monic polynomial of smallest degree in F[x] such that α is a root of f(x).

## Proof

(a) follows from the Corollary to Theorem 11.1.3.

(b) If f(x) is not of the smallest degree polynomial in F[x], of which α is a root, then there exists a monic polynomial g(x) in F[x] such that α is a root of g(x) and degg(x)<degf(x). This gives a contradiction. □

## Definition 11.1.8

Let L and K be two field extensions of the same field F. An isomorphism ψ:LK of fields, whose restriction to F is the identity homomorphism, is called an isomorphism of field extensions or an F-isomorphism. Two field extensions L and K of the same field F are said to be isomorphic field extensions iff there exists an F-isomorphism ψ:LK.

## Proposition 11.1.2

Let L and K be two field extensions of the same field F and ψ:LK be an F-isomorphism. Suppose f(x)∈F[x] and α is a root of f(x) in L. If β=ψ(α)∈K, then β is also a root of f(x).

## Proof

If f(x)=a 0+a 1 x+⋯+a n x n F[x], then ψ(a i )=a i and ψ(α)=β. Again f(α)=0 gives 0=ψ(0)=ψ(f(α))=ψ(a 0+a 1 α+⋯+a n α n )=ψ(a 0)+ψ(a 1)ψ(α)+⋯+ψ(a n )(ψ(α)) n , since ψ is a homomorphism. This shows that 0=a 0+a 1 β+⋯+a n β n . Hence β=ψ(α) is a root of f(x). □

## Theorem 11.1.5

Let F be a field and L,K be two field extensions of F. If αL and βK are both algebraic elements over F, then there is an isomorphism ψ:F(α)→F(β) of fields, such that ψ(α)=β and ψ| F = identity, iff α and β have the same minimal polynomial over F.

## Proof

Suppose that f(x) is the minimal polynomial of both α and β over F. Then by Theorem 11.1.3, there exist two isomorphisms μ:F[x]/〈f(x)〉→F(α) and λ:F[x]/〈f(x)〉→F(β). Hence the composite isomorphism ψ=λμ −1:F(α)→F(β) shows that F(α) and F(β) are isomorphic under ψ.

Conversely, suppose ψ:F(α)→F(β) is an isomorphism such that ψ(α)=β and ψ| F =identity. If f(x)∈F[x] is a polynomial such that f(α)=0, then ψ(α)=β is also a root of f(x) by Proposition 11.1.2. Hence α and β have the same minimal polynomial. □

## Example 11.1.6

If ω=e 2π/3 is a complex cube root of 1 and α is a real root of the polynomial f(x)=x 3−3 in Q[x], then f(x) is the minimal polynomial of both α and ωα over Q. Hence by Theorem 11.1.5 there exists a Q-isomorphism ψ:Q(α)→Q(ωα), such that ψ(α)=ωα. Note that the elements of Q(α) are real numbers, but the elements of Q(ωα) are not.

Let F be a field and K be a field extension of F. Then we may consider K as a vector space over F by using the field operations.

## Definition 11.1.9

Let F be a field and K a field extension of F. Then the dimension of the vector space K over F is called the degree or dimension of K over F and is denoted by [K:F]. If [K:F] is finite, then K is called a finite extension over F (or K is said to be finite over F) or K/F is called a finite extension. Otherwise, K is said to be an infinite extension of F.

## Example 11.1.7

(i) C is a finite extension over R. This is so because {1,i} forms a basis of C over R and hence [C:R]=2.

(ii) Let $$K=\mathbf{Q}( \{\sqrt{q}: q \mbox{ is a prime integer} \})\subset\mathbf{R}$$. Then [K:Q] is not finite.

[Hint. As q is a prime integer, $$\sqrt{q}$$ is not an element of Q. Let us assume that p 1,p 2,…,p n be distinct prime integers (each is different from q) such that $$\sqrt{q}\notin\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$$, which is our induction hypothesis. We claim that $$\sqrt{q}\notin\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}},\sqrt{p_{n+1}})$$, where p 1,p 2,…,p n ,p n+1 are distinct prime integers such that each is different from q. As $$\sqrt{q}\notin\mathbf{Q}$$, the induction hypothesis is true for n=0. If $$\sqrt{q}\in\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}},\sqrt{p_{n+1}})$$, then there exist elements $$x,y\in\mathbf{Q}(\sqrt{p_{1}},\ldots,\sqrt{p_{n}})$$ such that $$\sqrt{q}=x+y\sqrt{p_{n+1}}$$. As y=0 contradicts the induction hypothesis, it follows that y≠0. If x=0, then q=y 2 p n+1 gives also a contradiction, since q and p n+1 are distinct primes. Again for x≠0 and y≠0, $$q=x^{2}+2xy\sqrt{p_{n+1}}+p_{n+1}y^{2} \Rightarrow\sqrt{p_{n+1}}=(q-x^{2}-p_{n+1}y^{2})/2xy\in\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})\Rightarrow\sqrt{q}\in \mathbf{Q}(\sqrt{p_{1}}, \sqrt{p_{2}},\ldots,\sqrt{p_{n}})$$. This gives a contradiction of hypothesis again. Thus by the induction hypothesis, we find that for any positive integer r, if p 1,p 2,…,p r ,q are distinct prime integers, then $$\sqrt{q}\notin\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{r}})$$. Consequently, we get a strictly infinite ascending chain of fields such that $$\mathbf{Q}\subsetneqq\mathbf{Q}(\sqrt{2})\subsetneqq\mathbf{Q}(\sqrt{2},\sqrt{3}) \subsetneqq\cdots$$. This shows that [K:Q] is not finite. Clearly, [R:Q] is not finite.]

## Theorem 11.1.6

Let K be a finite field extension of the field F. If [K:F]=n, then
1. (a)

there exists a basis of K over F, consisting of n elements of K and

2. (b)

any set of (n+1) (or more) elements of K is always linearly dependent over F.

## Proof

As K is an n-dimensional vector space over F, the theorem follows from the properties of finite dimensional vector spaces. □

## Corollary 1

The degree of an algebraic element α over a field F is equal to the dimension of the extension field F(α), regarded as a vector space over F, with a basis B={1,α,α 2,…,α n−1} if [F(α):F]=n.

## Corollary 2

Let α and β be two algebraic elements over the same field F such that F(α)=F(β). Then α and β have the same degree over F.

## Theorem 11.1.7

Let K be a finite field extension of the field F and an element α of K be algebraic over F. If n is the degree of the minimal polynomial f(x) of α over F, then [F(α):F]=n.

## Proof

It is sufficient to prove that the set S={1,α,…,α n−1} forms a basis of the vector space F(α) over F. Suppose a 0+a 1 α+⋯+a n−1 α n−1=0, where each a i F. Then α is a root of the polynomial g(x)=a 0+a 1 x+⋯+a n−1 x n−1F[x]. Hence f(x) divides g(x) in F[x] by Theorem 11.1.4(a). But this is possible only when a 0=a 1=⋯=a n−1=0, since degg(x)<degf(x)=n. This concludes that S is linearly independent over F. We now claim that the set S generates the vector space F(α). Let h(α)∈F(α). As α is algebraic over F, F(α)=F[α] (see Proposition 11.1.1). Hence h(α)∈F[α]. Let h(x) be the polynomial in F[x] corresponding to h(α) in F[α]. Again F is a field ⇒F[x] is a Euclidean domain ⇒ there exist polynomials q(x) and r(x) in F[x] such that h(x)=f(x)q(x)+r(x), where r(x)=0 or degr(x)<degf(x). Hence for h(α)=f(α)g(α)+r(α), either h(α)=0 or h(α)=r(α) is a linear combination of elements of S, since degr(x)<n. Thus S forms a basis of F(α) over K and hence [F(α):F]=n. □

## Theorem 11.1.8

Let K be a finite field extension of the field F. If S={α 1,α 2,…,α n }⊂K forms a basis of the vector space K over F, then K=F(α 1,α 2,…,α n ).

## Proof

Clearly, F(α 1,α 2,…,α n )⊆K, since F(α 1,α 2,…,α n ) is the smallest field containing F and S. For the reverse inclusion, let αK. As S forms a basis of K, there exist elements a 1,a 2,…,a n in F such that α=a 1 α 1+a 2 α 2+⋯+a n α n F(α 1,α 2,…,α n ). Thus KF(α 1,α 2,…,α n ) and hence K=F(α 1,α 2,…,α n ). □

We now prove a relation between a finite field extension and an algebraic extension.

## Theorem 11.1.9

Every finite field extension of F is an algebraic extension.

## Proof

Let K be a finite field extension of the field F. Suppose [K:F]=n and α is a non-zero element of K. Then the (n+1) elements 1,α,α 2,…,α n of the vector space K over F are linearly dependent. Hence there exist elements a 0,a 1,…,a n (not all zero) in F such that a 0+a 1 α+⋯+a n α n =0. This shows that α is a root of the non-null polynomial a 0+a 1 x+⋯+a n x n F[x]. This implies α and hence K is algebraic over F. □

## Corollary

If K is a finite field extension of F, then every element of K is algebraic over F.

## Remark

The converse of Theorem 11.1.9 is not true in general. For example, the field K defined in Example 11.1.7(ii) is an algebraic field extension of Q but K is not a finite field extension of Q. This is so because for αK, there exist prime integers p 1,p 2,…,p n such that $$\alpha\in\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt {p_{n}})$$. But $$\mathbf{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$$ is a finite extension of Q implies that α is algebraic over Q. Thus K is algebraic over Q but [K:Q] is not finite. The partial converse of Theorem 11.1.9 is true for the following particular case.

## Theorem 11.1.10

The field extension K(α) over K is finite iff α is algebraic over K.

## Proof

If K(α) is a finite field extension over K, then by the Corollary to Theorem 11.1.9, it follows that α is algebraic over K. Conversely, let α be algebraic over K. Then K(α) is an extension over K implies that [K(α):K]=n, where n is the degree of the minimal polynomial f(x) of α over K. This proves that K(α) is a finite extension over K. □

## Corollary

Every element of a simple algebraic extension F(α) is algebraic over F.

## Remark

A transcendental element cannot appear in a simple algebraic extension.

## Definition 11.1.10

Let K be a field extension of the field F. A subfield L of K is said to be an intermediate field of K/F iff FLK holds and this intermediate field L is said to be proper iff LK and LF.

## Example 11.1.8

Let K be a field extension of a field F and α be an element of K. Then the field F(α) generated by F and α is an intermediate field, because FF(α)⊆K.

## Remark

Let L be an intermediate field of K/F. Then
1. (a)

K is a vector space over F and L is a subspace of K;

2. (b)

K/F is an algebraic extension iff K/L and L/F are both algebraic extensions.

## Theorem 11.1.11

Let K be a finite field extension of the field F and L be an intermediate field of K/F. Then [K:F]=[K:L][L:F].

## Proof

Let V be a basis of the vector space K over L and U be a basis of the vector space L over F. Suppose [L:F]=m and [K:L]=n. Let V={v 1,v 2,…,v n } and U={u 1,u 2,…,u m }. Construct W={uv:uU,vV}. Then W forms a basis of K over F. Finally, $$\operatorname{card}W=nm$$ proves the theorem. □

## Remark

If either $$\operatorname{card}V$$ or $$\operatorname{card}U$$ is infinite, then $$\operatorname{card}W$$ is also infinite.

## Corollary 1

If K is a field extension of the field F and each of the elements α 1,α 2,…,α n of K are algebraic over F, then K(α 1,α 2,…,α n ) is a finite extension of F.

## Proof

The corollary follows by induction on n. □

## Corollary 2

If V={v 1,v 2,…,v n } is a basis of a finite field extension K of L and U={u 1,u 2,…,u m } is a basis of a finite field extension L of F, then W={uv:uU,vV}, consisting of mn elements, forms a basis of K over F.

## Corollary 3

Let F be a field and K be a field extension of F such that [K:F]=n. If an element α of K is algebraic over F, then the degree of α over F divides n.

## Proof

Consider the tower of fields: FF(α)⊂K. Then n=[K:F]=[K:F(α)][F(α):F]. As α is algebraic over F, then [F(α):F] is the degree of α over F. Hence the corollary follows. □

## Corollary 4

Let F be a field and K be a field extension of F of prime degree p. If α is an element of K but not in F, then α has the degree p over F and K=F(α).

## Proof

[K:F]=p⇒[K:F(α)][F(α):F]=p. Again αF⇒[F(α):F]≠1⇒[K:F(α)]=1 and [F(α):F]=p. The corollary follows from Ex. 1 of Exercise-I. □

## Corollary 5

Let FLK be a tower of fields such that K is algebraic over L and L is algebraic over F. Then K is algebraic over F.

## Proof

It is sufficient to show that every element αK is algebraic over F. By hypothesis, K is algebraic over L⇒∃a 0,a 1,…,a n−1 (not all zero) ∈L such that α n +a n−1 α n−1+⋯+a 1 α+a 0=0⇒α is algebraic over the field F(a 0,a 1,…,a n−1). Consider the tower of fields: FF(a 0)⊂F(a 0,a 1)⊂⋯⊂F(a 0,a 1,…,a n−1)⊂F(a 0,a 1,…,a n−1,α). Clearly, each extension of the above tower is finite. Hence by Theorem 11.1.11, the degree of F(a 0,a 1,…,a n−1,α) over F is finite. Then it follows that α is algebraic over F. Hence K is algebraic over F. □

The structure of a simple transcendental extension is given in the next theorem.

## Theorem 11.1.12

If α is transcendental over a field F, then the field F(α) generated by F and α is isomorphic to the field F(x) of all rational functions in x over F.

## Proof

The extension F(α) is given by F(α)={f(α)/g(α):f(x),g(x)∈F[x] and g(α)≠0}, using Corollary to Theorem 11.1.1. If two polynomial expressions f 1(α) and f 2(α) are equal in F(α), their coefficients must be equal term by term, otherwise, the difference f 1(α)−f 2(α) will give a polynomial equation in α with coefficients, not all zero. This contradicts our assumption that α is transcendental over F. This shows that the map ψ:F[α]→F[x], f(α)↦f(x) is a bijection and hence it is an isomorphism under usual operations of polynomials. This isomorphism gives rise to an isomorphism F(α)→F(x), f(α)/g(α)↦f(x)/g(x). □

## Exercises-I

1. 1.

Let K be a field extension of the field F. Prove that [K:F]=1 iff K=F.

[Hint. [K:F]=1⇒K is a vector space over F of dimension 1 with {1} as a basis. Now xKx=a.1 for some aFKF. Then FK and KFK=F. Conversely, if K=F, then {1} is a basis of the vector space K over F. Hence [K:F]=1.]

2. 2.

Let K be a field extension of the field F such that [K:F] is finite. If f(x) is an irreducible polynomial in F[x] such that f(α)=0 for some αK, then prove that degf(x) divides [K:F].

[Hint. Suppose degf(x)=n. Then [F(α):F]=n⇒[K:F]=[K:F(α)][F(α):F]=[K:F(α)]⋅n.]

3. 3.

Find $$[\mathbf{Q} (\sqrt[3]{5}): \mathbf{Q}]$$.

[Hint. $$\sqrt[3]{5}$$ has the minimal polynomial x 3−5 over Q. Hence $$[\mathbf{Q} (\sqrt[3]{5}): \mathbf{Q}]= 3$$.]

4. 4.

Let K be a field extension of the field F and a,bK be algebraic over F. If a has degree m over F and b≠0 has degree n over F, then show that the elements a+b, ab, ab and ab −1 are algebraic over F and each has at most degree mn over F.

5. 5.

Let K be a field extension of F such that [K:F]=p, where p is a prime integer. Then show that K/F has no proper intermediate field.

[Hint. Suppose K/F has an intermediate field L. Then [K:F]=[K:L][L:F]⇒[L:F] divides p⇒ either [L:F]=1 or [L:F]=p⇒ either L=F or L=K.]

6. 6.

Let K be a field extension of F and L be the set of all elements of K which are algebraic over F. Then show that L is an intermediate field of K/F.

7. 7.

Let K be a field extension of F and L be an intermediate field of K/F. Show that L is an algebraic extension iff both K/F and L/F are algebraic extensions.

8. 8.

Let K be a field extension of the field F and R be a ring such that FRK. If every element of R is algebraic over F, then prove that R is a field.

[Hint. Let a be a non-zero element of R and f(x)=x n +⋯+a 1 x+a 0 be its minimal polynomial over F. Then a n +⋯+a 1 a+a 0=0. If a 0=0, we have a contradiction, since the minimal polynomial of α over F has degree n. Again if a 0≠0, then a −1 exists.]

9. 9.

Show that $$\mathbf{Q}(\sqrt{5},-\sqrt{5})=\mathbf{Q}(\sqrt{5})$$.

[Hint. $$\mathbf{Q}(\sqrt{5},-\sqrt{5})$$ is the smallest field containing $$\sqrt{5},- \sqrt{5}$$ and Q. Again $$\sqrt{5}\in\mathbf{Q}(\sqrt{5})\Rightarrow-\sqrt{5} \in\mathbf{Q}(\sqrt{5})$$. Moreover, $$\mathbf{Q}\subset\mathbf{Q}(\sqrt{5}), \mathbf{Q}(\sqrt{5},-\sqrt{5}) \subseteq\mathbf{Q}(\sqrt{5})$$ and $$\sqrt{5} \in\mathbf{Q}(\sqrt{5},-\sqrt{5}) \Rightarrow$$ $$\mathbf{Q}(\sqrt{5})$$ is the smallest field containing $$\mathbf{Q}, \sqrt{5}$$ and $$- \sqrt{5}$$.]

10. 10.

Show that x 2−5 is irreducible over $$\mathbf{Q}(\sqrt{2})$$.

[Hint. Suppose x 2−5 is not irreducible over $$\mathbf{Q}(\sqrt{2})$$. Then x 2−5=(xa)(xb) for some $$a,b \in\mathbf{Q}(\sqrt{2})$$. Hence a+b=0 and $$ab=-5 \Rightarrow\sqrt{5} \in\mathbf{Q}(\sqrt{2})$$. This is a contradiction.]

11. 11.

Let K be finite extension of the field F. If α,βK are algebraic over F, then show that α+β, αβ and α −1 (α≠0) are also algebraic over F.

[Hint. Clearly, F(α,β) is a finite extension of F shows that F(α,β) is an algebraic extension of F.]

12. 12.
(a) If α is transcendental over a field F, then show that
1. (i)

the map μ:F[x]→F[α],f(x)↦f(α) is an isomorphism;

2. (ii)

F(α) is isomorphic to the field F(x) of rational functions over F in the indeterminate x.

(b) Show that the field Q(π) is isomorphic to the field Q(x) of rational functions over Q.

[Hint. See Theorem 11.1.12.]

13. 13.

For the field extension R of Q, show that π is transcendental over $$\mathbf{Q}(\sqrt{2})$$.

14. 14.

Let K be a field extension of F and L be also a field extension of F. If αK and βL are both algebraic over F, then show that there is an isomorphism of fields: μ:F(α)→F(β), which is the identity on the subfield F and which maps αβ iff the monic irreducible polynomials for α and β over F are the same.

15. 15.

If f(x) is an irreducible polynomial of degree n over a field F, then show that there is an extension K of F such that [K:F]=n and f(x) has a root in K.

16. 16.

Let K be a field extension of the field F and α,β be two algebraic elements of K over F having degrees m and n respectively. If gcd(m,n)=1, show that [F(α,β):F]=mn.

17. 17.

Show that every irreducible polynomial in R[x] has degree 1 or 2.

[Hint. Let f(x) be an irreducible polynomial in R[x]. Then f(x) has a root in C by s of Algebra. Since [C:R]=2, the degree of α over R divides 2.]

## 11.2 Finite Fields

Given a prime integer p and a positive integer n, we have shown in Chap. the existence of a finite field with p n elements.

Finite fields form an important class of fields. A field having finitely many elements is called a finite field. If F is a finite field, the kernel of the homomorphism ψ:ZF, nn1 F is a prime ideal. Then kerψ≠{0}, because F is finite but Z is infinite. Thus kerψ=〈p〉, generated by a prime integer p and ψ(Z) is isomorphic to the quotient field Z/〈p〉=Z p . So F contains a subfield K isomorphic to the prime field Z p . Hence F can be considered as an extension of Z p .

In this section we study finite fields. Throughout the section p denotes a prime integer.

## Definition 11.2.1

A field having finitely many elements is called a finite field or Galois field.

Let K be a finite field. If $$\operatorname{char} K=p$$, then by Example 11.1.1(iv), K may be considered as a finite dimensional vector space over Z p . Then the dimension of K over Z p is finite and it is denoted by [K:Z p ].

## Theorem 11.2.1

Let K be a finite field of characteristic p and [K:Z p ]=n. Then K contains exactly p n elements.

## Proof

Let B={x 1,x 2,…,x n } be a basis of K over Z p and xK. Then x can be expressed as x=a 1 x 1+a 2 x 2+⋯+a n x n , where a i Z p . As Z p has p elements, K has at most p n elements. Since B is a basis of K, the elements a 1 x 1+a 2 x 2+⋯+a n x n are all distinct for every distinct choice of elements a 1,a 2,…,a n of Z p . Thus K has exactly p n elements. □

## Theorem 11.2.2

Let K be a finite field of characteristic p. Then every element of K is a root of the polynomial $$x^{p^{n}}-x$$ over Z p .

## Proof

Let [K:Z p ]=n. Then K is a finite field of p n elements. Hence the multiplicative group K∖{0} is of order p n −1. If y is a non-zero element of K, then $$y^{p^{n}-1}=1$$ and hence $$y^{p^{n}}=y$$. Moreover, $$0^{p^{n}}=0$$. Thus every element of K is a root of the polynomial $$x^{p^{n}}-x$$ over Z p . □

To describe finite fields, we need the concept of splitting fields.

## Definition 11.2.2

Let F be a field. A polynomial f(x) in F[x] is said to split over a field K containing F iff f(x) can be factored as a product of linear factors in K[x]. A field K containing F is said to be a splitting field for f(x) over F iff f(x) splits over K and there is no proper intermediate field L of K/F (i.e., if L is a subfield of K such that FLK, then either L=F or L=K).

## Remark 1

Let F be a field and f(x) be a polynomial in F[x] of positive degree n. Then an extension K of F is a splitting field of f(x) such that
1. (i)

f(x) factors into linear factors in K[x] as f(x)=a(xα 1)(xα 2)⋯(xα n ), with α i K and aF;

2. (ii)

K is generated by F and the roots of f(x), i.e., K=F(α 1,α 2,…,α n ).

The second condition shows that K is the smallest extension of F which contains all the roots of f(x).

## Remark 2

If F is a field, then every polynomial f(x) in F[x] of positive degree has a splitting field over F.

## Remark 3

Let F be a field and f(x)∈F[x] is of positive degree. Then any two splitting fields of f(x) are isomorphic and hence the splitting field of f(x) is unique up to isomorphism.

## Example 11.2.1

(i) C is the splitting field of the polynomial x 2+1 over R but R is not splitting field of x 2+1 over Q.

(ii) Let K be a finite field of characteristic p. Then K is the splitting field of $$x^{p^{n}}-x$$ over Z p . This is so because if S is the splitting field of $$x^{p^{n}}-x$$ over Z p , then S contains all the roots of the polynomial $$x^{p^{n}}-x$$ over Z p . Hence SK. Again by Theorem 11.2.2, KS. Consequently K=S.

## Exercises-II

1. 1.

Show that any two finite fields with same number of elements are isomorphic.

[Hint. Let K and S be two finite fields containing p n elements, where p is prime and n is a positive integer. Then K and S are both the splitting fields of the same polynomial $$x^{p^{n}}-x$$ over Z p and hence they are isomorphic.]

2. 2.

Corresponding to a given prime integer p, and a positive integer n, show that there exists a finite field consisting of p n =q roots of x q x over Z p , which is determined uniquely up to an isomorphism and is denoted by $$\mathbf{F}_{p^{n}}$$. This field is sometimes called the Galois field GF(p n ).

3. 3.

Prove that $$\mathbf{F}_{p^{n}}$$ is a subfield of $$\mathbf{F}_{p^{m}}$$ iff n is a divisor of m.

[Hint. Let H be a subfield of $$\mathbf{F}_{p^{m}}$$. Then H is a finite extension of Z p and $$H= \mathbf{F}_{p^{n}}$$, where n=[H:Z p ]. Thus $$m=[\mathbf{F}_{p^{m}}: \mathbf{Z}_{p}]=[\mathbf{F}_{p^{m}}: H][H: \mathbf{Z}_{p}]$$. Conversely, if n is a divisor of m, then (p n −1)|(p m −1) and hence $$(x^{p^{n}}-x) | (x^{p^{m}} -x)$$. Consequently, all the roots of $$x^{p^{n}}-x$$, over Z p , are contained among the roots of $$x^{p^{m}}-x$$ over Z p .]

4. 4.

Let F be a field and G be a finite multiplicative subgroup of F =F∖{0}. Then show that G is cyclic and G consists of all the nth roots of unity in F, where |G|=n. (A generator of G is called a primitive element for F.)

[Hint. Use the Structure Theorem for abelian groups or see Chap. .]

5. 5.

Let F be a finite field. For the finite extension F(a,b) of F with a,b algebraic over F, show that there exists an element c in F(a,b) such that F(a,b)=F(c), i.e., F(a,b) is a simple extension of F.

6. 6.

Let F be a field of p n elements, where p is prime and n is a positive integer. Then show that [F:Z p ]=n.

[Hint. Suppose [F:Z p ]=m. Then by Theorem 11.2.1, F has exactly p m elements. Hence p m =p n m=n.]

7. 7.
Let F be a finite field of characteristic p. Then prove that
1. (a)

if c is a primitive element for F, then c p is also;

2. (b)

if n=[F:Z p ], then there exists an element c in F such that c is algebraic over Z p , of degree n and F=Z p (c).

8. 8.

Every finite field F of characteristic p has an automorphism ψ:FF,xx p . Find the order of ψ in the automorphism group of F. If F is not finite, examine the validity of the above result.

[Hint. For the field F=Z p (x), the field of rational functions, the map ψ:FF,xx p is not an automorphism.]

9. 9.

Let F be a finite field and K a finite field extension of F. Then prove that the number of elements of K is some power of the number of elements of F.

[Hint. If $$\operatorname{char} F = p >0$$, then F has p m =q (say) elements. Again if [K:F]=n, then K has q n elements.]

## 11.3 Algebraic Numbers

An algebraic number is a complex number which is algebraic over the field Q of rational numbers. In this section we shall discuss the basic properties of algebraic numbers. The theory of algebraic numbers was born and developed through the attempts to prove Fermat’s Last Theorem. In this section we study the properties of algebraic numbers, such as the set $$\mathcal{A}$$ of algebraic numbers is countable and forms a field, called the algebraic number field which is algebraically closed. We prove the fundamental theorem of algebra and show the existence of real transcendental numbers.

## Definition 11.3.1

A complex number α which is algebraic over Q (i.e., α is a root of some non-null polynomial over Q) is called an algebraic number. Otherwise, it is said to be transcendental.

## Example 11.3.1

(i) Every rational number q is an algebraic number. This is so because q satisfies the equation xq=0 over Q.

(ii) $$1/\sqrt{5}$$ is an algebraic number. This is so because it satisfies the equation x 2−1/5=0 over Q.

(iii) e and π are transcendental numbers [see Hardy and Wright (2008, pp. 218–227)].

## Theorem 11.3.1

An algebraic number α satisfies a unique monic irreducible polynomial equation f(x)=0 over Q. Moreover, every polynomial equation g(x)=0 over Q satisfied by α is divisible by f(x) in Q[x].

## Proof

Let S be the set of all polynomial equations over Q satisfied by the given algebraic number α and h(x)=0 be one of the lowest degree polynomial equations in S. If the leading coefficient of h(x) is q, define f(x) by f(x)=q −1 h(x). Hence f(α)=0 and f(x) is monic. We claim that f(x) is irreducible in Q[x]. Suppose f(x)=f 1(x)f 2(x) in Q[x]. Then at least one of f 1(α)=0 and f 2(α)=0 holds. This implies a contradiction, because f(x)=0 is a polynomial equation of lowest degree in S having α as a root. Again, since Q is a field, Q[x] is Euclidean domain. Hence for the polynomials f(x) and g(x), there exist polynomials q(x) and r(x) in Q[x] such that g(x)=f(x)q(x)+r(x), where r(x)=0 or degr(x)<degf(x). Then r(x) must be identically zero, otherwise, the degree of r(x) would be less than the degree of f(x) and α would be a root of the polynomial equation r(x)=0. Consequently, f(x) divides g(x) in Q[x]. To prove the uniqueness of f(x), suppose m(x) is an irreducible monic polynomial over Q such that m(α)=0. Then f(x) divides m(x). Hence there exists some polynomial n(x) in Q[x] such that m(x)=f(x)n(x). Since m(x) is irreducible in Q[x], n(x) must be constant in Q[x]. Again, since f(x) and m(x) are both monic polynomials, it follows that n(x)=1. This proves the uniqueness of f(x). □

This theorem leads to the following definition.

## Definition 11.3.2

Let α be an algebraic number. Then the irreducible monic polynomial f(x) over Q having α as a root is called the minimal polynomial of α and the degree of f(x) is called the degree of α.

## Example 11.3.2

$$\sqrt{3}$$ is an algebraic number and f(x)=x 2−3 is the minimal polynomial of $$\sqrt{3}$$ over Q. The field extension $$\mathbf{Q}(\sqrt{3})$$ of Q is of degree 2 with a basis $$B=\{1,\sqrt{3}\}$$.

## Theorem 11.3.2

Let F be a finite field extension of Q of degree n. Then every element α of F is an algebraic number and is a root of an irreducible polynomial over Q, of degreen.

## Proof

Clearly, the set S={1,α,α 2,…,α n−1,α n }, consisting of n+1 elements of the n-dimensional vector space F over Q, is linearly dependent over Q. This implies that α is algebraic over Q and hence α is an algebraic number. □

## Corollary

Every element of a simple algebraic extension Q(α) is an algebraic number.

## Definition 11.3.3

A field F is said to be algebraically closed (or complete) iff every polynomial in F[x], with degree ≥1, has a root in F.

Two important algebraically closed fields, such as the field C of complex numbers and the field of algebraic numbers are discussed here.

First we show that C is algebraically closed by the Fundamental Theorem of Algebra. There are several proofs of the Fundamental Theorem of Algebra. We present one of them by using the concept of homotopy as discussed in Sect.  of Chap. .

## Theorem 11.3.3

(Fundamental Theorem of Algebra)

Every non-constant polynomial with complex coefficients has a complex root.

## Proof

To prove the theorem it is sufficient to prove that a non-constant polynomial
$$f(z)=z^n+a_{n-1}z^{n-1}+ \cdots+ a_1z+a_0,\quad a_i \in\mathbf{C}$$
(11.1)
has a root in C. If a 0=0, then 0 is a root of f(z). So we assume that a 0≠0. We may consider f as a continuous function f:CC, defined by (11.1). Suppose f(z) has no root in C. Then f(z) is never zero. Consider the unit circle S 1={zC:|z|=1} in the complex plane. Define a continuous family of maps f t :S 1S 1,zf(z)/|f(tz)| for non-negative real numbers t. Then any two maps of this family are homotopic. For t=0, f 0 is a constant map. On the other hand, for sufficiently large t,f t is homotopic to g. where g(z)=z n , because z n is dominant in the expression of f(z) for sufficiently large t. Hence f 0 cannot be homotopic to g, because their degrees are different. This contradiction concludes that f(z) has a root in C. □

## Corollary 1

The field C of complex numbers is algebraically closed.

## Remark

This corollary proves the algebraic completeness of the field of complex numbers.

## Corollary 2

The field R of real numbers is embedded in the algebraically closed field C of complex numbers.

We now show that the set of algebraic numbers forms an algebraically closed field.

## Theorem 11.3.4

(a) The set $$\mathcal{A}$$ of algebraic numbers is a field, called the algebraic number field.

(b) The field $$\mathcal{A}$$ is algebraically closed.

## Proof

(a) It is sufficient to show that the sum, product, difference and quotient of any two elements α and β≠0 in $$\mathcal{A}$$ are also in $$\mathcal{A}$$. Clearly, α+β, αβ, αβ and αβ −1 are in the subfield Q(α,β) of C, which is generated by Q and two elements α and β of $$\mathcal{A}$$. Again α is algebraic over QQ(α) is a finite extension of Q and similarly β is algebraic over Q(α)⇒Q(α,β) is a finite extension over Q(α). Hence Q(α,β) is a finite extension of Q⇒ each element of Q(α,β) is an algebraic number. This shows that $$\mathcal{A}$$ is a field.

(b) Let f(x)=x n +a n−1 x n−1+⋯+a 0 be a polynomial in $$\mathcal{A}[x]$$. Then the coefficients a 0,a 1,…,a n−1 generate an extension F=Q(a 0,a 1,…,a n−1). Clearly, F is a finite extension of the field Q. Any complex root α of f(x) is algebraic over the field F and hence F(α) is a finite extension of F. Consequently, F(α) is also a finite extension of Q. Thus the element α of this extension is an algebraic number, in the field $$\mathcal{A}$$. This leads us to conclude that the field $$\mathcal{A}$$ is algebraically closed. □

## Corollary

The field Q of rational numbers is embedded in the field $$\mathcal{A}$$ of algebraically closed field $$\mathcal{A}$$.

We now show the existence of transcendental numbers. It does not follow immediately that there are transcendental numbers, though almost all real numbers are transcendental. So we need an alternative definition of an algebraic number.

An algebraic number is a complex number α which satisfies an algebraic equation of the form a 0 x n +a 1 x n−1+⋯+a n =0, where a 0,a 1,…,a n are all integers, not all zero. A number which is not an algebraic number is called a transcendental number.

## Theorem 11.3.5

The set of all algebraic numbers is countable.

## Proof

To prove this theorem we define the rank r of an equation $$a_{0}x^{n}+a_{1}^{n-1}+ \cdots+a_{n}=0$$, a i Z and a 0≠0 as r=n+|a 0|+|a 1|+⋯+|a n |. The minimum value of rank r is 2. As there exist only a finite number of such equations of rank r, we may write them as:
$$A_{r,1}, A_{r,2},\ldots,A_{r,m_r}.$$
Arranging the equations in the sequence:
$$A_{2,1},A_{2,2},\ldots,A_{2,r_2}, A_{3,1},A_{3,2}, \ldots,A_{3,r_3}, A_{4,1},\ldots,$$
we can assign to them the integers 1,2,3,… . This shows that the aggregate of these equations is countable. Clearly, every algebraic number corresponds to at least one of these equations, and the number of algebraic numbers corresponding to at least one of these equations is finite. Hence the theorem follows. □

## Corollary

The set of all real algebraic numbers is countable and has measure zero.

## Existence of Transcendental Numbers

The set of all real numbers is not countable. On the other hand, the set of all real algebraic numbers is countable. Hence there exist real numbers which are not algebraic. This shows the existence of real transcendental numbers. This leads us to conclude the following.

## Theorem 11.3.6

Almost all real numbers are transcendental.

## Exercises-III

1. 1.

Show that the degree of an algebraic number α is equal to the dimension n=[Q(α):Q], with a basis B={1,α,…,α n−1}. (Here Q(α) is regarded as a vector space over Q.)

2. 2.

If two algebraic numbers α and β generate the same extension fields Q(α) and Q(β), i.e., Q(α)=Q(β), then show that α and β have the same degree.

3. 3.
Let F be a field. Show that the following statements are equivalent:
1. (a)

F is algebraically closed;

2. (b)

every irreducible polynomial in F[x] is of degree 1;

3. (c)

if a polynomial f(x)∈F[x] is of degree ≥1, then f(x) can be expressed as a product of linear factors in F[x];

4. (d)

if K is an algebraic field extension of F, then F=K.

4. 4.

Let F be a finite field. Then show that F cannot be algebraically closed.

We now introduce the concept of algebraic integers.

## 11.5 Algebraic Integers

The concept of algebraic integers is one of the most important discoveries in number theory. The ring of algebraic integers shares some properties of the ring of integers but it differs as regards many other properties.

## Definition 11.5.1

A complex number α is called an algebraic integer iff α is a root of a monic irreducible polynomial f(x) in Q[x] with integral coefficients of the form
$$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0.$$

## Remark 1

An algebraic number α is an algebraic integer iff α satisfies some monic polynomial equation f(x)=0 with integral coefficients.

## Remark 2

While defining algebraic integers, some authors do not insert the condition of irreducibility of f(x) in Q[x], because of Theorem 11.5.2. The irreducible equation satisfied by a rational number p/q is just the linear equation xp/q=0. Thus a rational number is an algebraic integer iff it is an integer in the ordinary sense. Such an integer in Z is called a rational integer to distinguish it from other algebraic integers.

## Theorem 11.5.1

Every algebraic number is of the form α/β, where α is an algebraic integer and β is a non-zero integer in Z.

## Proof

Let γ be an algebraic number. Then γ is a root of a monic irreducible polynomial f(x)=x n +a n−1 x n−1+⋯+a 1 x+a 0Q[x]. If β is the lcm of the denominators of a 0,a 1,…,a n−1, then β is a positive integer and each βa i is an integer, for i=1,2,…,n−1. Then βγ is a root of a monic polynomial in Z[x]. Hence βγ is an algebraic integer, α (say). Thus γ=α/β, where α is an algebraic integer and β is a non-zero integer. □

## Definition 11.5.2

An algebraic element α≠0 is called a unit iff both α and α −1 are algebraic integers.

## Example 11.5.1

(i) $$\sqrt{3}$$ is an algebraic integer, because it satisfies the equation x 2−3=0 over Z.

(ii) $$(1+\sqrt{13})/2$$ is an algebraic integer, because it satisfies the equation x 2x−3=0 over Z.

(iii) $$\omega=\frac{1}{2}(-1+\sqrt{3}i)$$ is an algebraic integer, because it satisfies the equation x 2+x+1=0 over Z. In general, every root of unity is an algebraic integer, because it satisfies the monic polynomial equation x n −1=0 over Z for some positive integer n.

(iv) 0,±1,±2,… are the only algebraic integers in Q (referred to as “rational integers”).

[Hint. Every integer n is an algebraic integer, because n is a root of (xn)∈Z[x]. On the other hand, suppose a rational number m/q with gcd(m,q)=1 is an algebraic integer. Then there exists a polynomial f(x)=x n +a n−1 x n−1+⋯+a 0Q[x] such that f(m/q)=0 Consequently, m n +a n−1 qm n−1+⋯+a 0 q n =0⇒q|m n q=±1, since gcd(m,q)=1⇒m/q is an integer.]

(v) $$1/ \sqrt{2}$$ is an algebraic number but it is not an algebraic integer.

[Hint. As $$1/ \sqrt{2}$$ satisfies the equation x 2−1/2 over Q. So it is an algebraic number. If possible, let $$1/\sqrt{2}$$ be an algebraic integer. Then it satisfies a monic polynomial f(x) over Z such that $$f(1/\sqrt{2})=(1/\sqrt{2})^{n}+a_{n-1}(1/\sqrt{2})^{n-1}+a_{n-2}(1/\sqrt{2})^{n-2}+\cdots+ a_{0}=0$$. Then $$(1+2a_{n-2}+4a_{n-4}+\cdots)+\sqrt{2}(a_{n-1}+2a_{n-3}+\cdots)=0$$. If a=1+2a n−2+4a n−4+⋯, and b=a n−1+2a n−3+⋯, then $$a+b\sqrt{2}=0$$. Again if b≠0, then $$\sqrt{2}=-a/b$$ is the quotient of two integers. This gives a contradiction. Hence b=0 and thus $$a+b\sqrt{2}=0$$ gives a=0. This also gives a contradiction, since a is odd.]

## Remark

In examining whether a given algebraic number is an algebraic integer, it is not necessary to appeal to an irreducible polynomial equation. This follows from Theorem 11.5.2.

## Theorem 11.5.2

A complex number α is an algebraic integer iff it satisfies over Q a monic polynomial equation with integral coefficients.

## Proof

If α is an algebraic integer, then clearly α satisfies a monic polynomial p(x) over Q with integral coefficients. Conversely, let a complex number α satisfy a monic polynomial equation over Q with integral coefficients. Then α also satisfies an irreducible polynomial, say f(x) over Q with integral coefficients. Note that any common divisor of these integral coefficients may be removed. So without loss of generality, we may assume that the gcd of these integral coefficient is 1, resulting f(x) to be a primitive polynomial in Z[x]. The given polynomial p(x) is monic and hence also primitive in Z[x]. Now, the polynomial p(x) is divisible in Q[x] by the irreducible polynomial f(x). Let p(x)=f(x)g(x), where g(x)∈Q[x]. Since both p(x) and f(x) are primitive polynomials in Z[x], it follows that g(x)∈Z[x]. Thus the leading coefficient 1 in p(x) is the product of the leading coefficients in f(x) and g(x). Hence, ±f(x) is monic, resulting that α is an algebraic integer by definition. □

## 11.6 Gaussian Integers

We recall that Z[i]={a+bi:a,bZ} forms a ring, called Gaussian ring, named after C.F. Gauss. The elements of Z[i], called Gaussian integers are the points of a square lattice in the complex plane. In this section we define Gaussian integers and Gaussian prime integers. Moreover, we study their properties with an eye to the corresponding properties of integers.

## Definition 11.6.1

A complex number of the form a+bi, a,bZ (i.e., an element of Z[i]), is called a Gaussian integer.

The set of all Gaussian integers is a subring of C. C.F. Gauss developed the properties of Gaussian integers in his work on biquadratic reciprocity. Z[i] is an Euclidean domain with norm function defined by $$N(\alpha)=\alpha\bar{\alpha}=m^{2}+n^{2}$$, for α=m+niZ[i]. Clearly, N(α) has the following properties in Z[i]:
1. (a)

N(αβ)=N(α)N(β) for all α,βZ[i]∖{0};

2. (b)

N(α)=1 iff α is a unit;

3. (c)

1, −1, i and −i are the only units in Z[i];

4. (d)

$$N(\alpha)=\left \{ \begin{array}{l@{\quad }l} 1 & \mbox{if } \alpha \mbox{ is a unit}\\ {>}1 & \mbox{if } \alpha\notin\{0,1,-1,i,-i\}. \end{array} \right .$$

## Definition 11.6.2

A prime element in the Gaussian ring Z[i] is called a Gaussian prime integer.

## Remark

Every Gaussian prime is non-zero and non-unit.

## Example 11.6.1

(i) 3 is a Gaussian prime integer.

(ii) 5 is not a Gaussian prime integer.

[Hint. As there are four units in Z[i], there are four factorizations of the integer 5 in Z[i], which are considered equivalent: 5=(2+i)(2−i). This is the prime factorization of 5 in Z[i]. Every non-zero element αZ[i] has four associates, viz., ±α. For example, the associates of 2+i are 2+i, −2−i, −1+2i, 1−2i.]

## Example 11.6.2

3+2i is an algebraic integer, because it is a root of x 2−6x+13=0.

## Proposition 11.6.1

For an element α in Z[i], if N(α)=p, a prime integer, then α is prime in Z[i].

## Proof

Clearly, α is a non-zero non-unit, since p≠0, p≠1 and 1, −1, i and −i are the only units in Z[i]. If possible, let α=βγ for some β,γZ[i]. Then N(α)=N(β)N(γ)⇒p=N(β)N(γ)⇒ either N(β)=1 or N(γ)=1, since p is prime and N(β),N(γ) are both positive integers. Hence either γ or β is a unit in Z[i]. This shows that α is irreducible in Z[i]. As Z[i] is an Euclidean domain, Z[i] is also a principal ideal domain. Hence α is a prime element in Z[i]. □

## Remark

N(3)=9⇒ norm of a Gaussian prime integer may not be a prime integer.

## Example 11.6.3

The Gaussian integer 1+i is Gaussian prime, since N(1+i)=2 is a prime integer.

## Remark

All prime integers are not Gaussian primes. Let p be a prime integer. Then p is a Gaussian prime iff the ring Z[i]/〈p〉 is a field.

We now describe all Gaussian primes.

## Theorem 11.6.1

The Gaussian primes are precisely of the following types:
1. (a)

all prime integers of the form 4n+3 and their associates in Z[i];

2. (b)

the number 1+i and its associates in Z[i];

3. (c)

all Gaussian integers α associated with either a+bi or abi, where a>0, b>0, one of a or b is even and N(α)=a 2+b 2 is a prime integer of the form 4m+1.

## Proof

Let α=a+bi be a prime element in Z[i]. Then N(α)=a 2+b 2 is an integer >1. Suppose $$N(\alpha) =\alpha\bar{\alpha}=p_{1}^{\alpha_{1}}p_{2} ^{\alpha_{2}}\cdots p_{m}^{\alpha_{m}}$$, where p 1,p 2,…,p m are distinct prime integers. Then $$\alpha| p_{1}^{\alpha_{1}}p_{2} ^{\alpha_{2}}\cdots p_{m}^{\alpha_{m}}$$ in Z[i]. We claim that α divides only one of the integers p 1,p 2,…,p m . If possible, let α divide two distinct primes p and q in the above expression. Now gcd(p,q)=1⇒1=px+qy for some x,yZα|1⇒ a contradiction, since α is a prime element in Z[i]. This shows that α divides only one prime p (say) in Z[i]. On dividing p by 4, we have either $$p\equiv 1\ (\operatorname{mod}4)$$ or $$p\equiv2\ (\operatorname{mod}4)$$ or $$p\equiv3\ (\operatorname{mod}4)$$.
1. (a)

Suppose $$p\equiv3\ (\operatorname{mod}4)$$. Now α|p in Z[i]⇒N(α)|N(p)=p 2N(α)=p or p 2, since N(α)>1. If N(α)=p, then a 2+b 2=p. Since p is an odd prime, either a is even and b is odd or a is odd and b is even which imply $$p\equiv 1\ (\operatorname{mod}4)\Rightarrow$$ a contradiction. Thus, N(α)=p 2. Again α|p in Z[i]⇒αβ=p for some βZ[i]⇒N(α)N(β)=N(p)⇒p 2 N(β)=p 2N(β)=1⇒β is a unit in Z[i]⇒α and p are associates in Z[i]. Thus p is also a prime element in Z[i], as required in (a).

2. (b)

Suppose $$p\equiv2\ (\operatorname{mod}4)$$. Then p=2=(1+i)(1−i). Now N(1+i)=2⇒1+i is prime in Z[i]. Again 1−i=−i(1+i) shows that 1−i is an associate of (1+i). Thus 1+i and its associates are prime elements in Z[i], as required in (b).

3. (c)

Suppose $$p\equiv1\ (\operatorname{mod}4)$$. Now α|pN(α)|N(p)⇒N(α)=p or p 2 . Suppose N(α)=p 2. As $$p\equiv1\ (\operatorname{mod}4)$$, suppose p=4k+1, kZ. Then the quadratic reciprocity of −1 modulo p, i.e., $${{-1}\choose{p}}= (-1)^{\frac{p-1}{2}}(\operatorname{mod}p)=(-1)^{2k}=1$$. Hence there exists an integer n such that $$n^{2}\equiv-1\ (\operatorname{mod}p)\Rightarrow p | (n^{2}+1)$$ in Zp|(n 2+1) in Z[i]⇒p|(n+i)(ni) in Z[i]. But α|pα|(n+i)(ni)⇒α|(ni) or α|(ni), since α is a Gaussian prime. We claim that p is not an associate of α. Suppose p is an associate of α. Then p|α in Z[i]⇒p|(n+i) or p|(ni) in Z[i]⇒ either $$\frac{n}{p}+\frac{1}{p}i$$ or $$\frac{n}{p}-\frac{1}{p}i$$ is a Gaussian integer ⇒ a contradiction, since 1/p is not an integer ⇒p is not an associate of α. Suppose N(α)=N(p)=p 2. Then α|pp=αβ for some βZ[i]. Hence N(p)=N(α)N(β)⇒N(β)=1⇒β is a unit in Z[i]⇒p is an associate of α⇒ a contradiction ⇒N(α)≠p 2N(α)=p (which is the other possibility) $$\Rightarrow N(\bar{\alpha})=p\Rightarrow a-bi$$ is also a Gaussian prime integer. Suppose abi is an associate of a+bi. Then abi=u(a+bi), where u∈{1,−1,i,−i}. If u=1, then b=0. Hence a 2=N(α)=p is not possible. Thus u≠1. Similarly, u≠−1,i,−i. Consequently, a+bi is not an associate of abi. Again N(α)=pa 2+b 2=p⇒ one of a and b must be even and the other must be odd. This proves (c).

□

Like division algorithm for ordinary integers and for polynomials, a division algorithm can be developed for Gaussian integers.

## Theorem 11.6.2

For given Gaussian integers α and β≠0, there exist Gaussian integers γ and λ in Z[i] such that
$$\alpha=\beta\gamma+\lambda,\quad \textit{where either}\ \lambda=0\ \textit{or}\ \quad N(\lambda)<N(\beta).$$
(11.2)

## Proof

Suppose α/β=r+ti and choose integers r′ and t′ as close as possible to the rational numbers r and t, respectively. Then α/β=(r′+ti)+[(rr′)+(tt′)i]=γ+μ (say), where |rr′|≤1/2, |tt′|≤1/2. Now α=βγ+βμ. As α, β and γZ[i], βμ=αβγZ[i]. If r=r′ and t=t′, then μ=0. If not, then μ≠0. Let β=a+biZ[i]. Then N(βμ)=(a 2+b 2)[(rr′)2+(tt′)2]<(a 2+b 2)=N(β). Thus the proposition follows by assuming λ=βμ. □

## Proposition 11.6.2

Two Gaussian integers α and β, not both zero, have a greatest common divisor δ in Z[i], which is a Gaussian integer given by δ=λα+μβ, where λ and μ are Gaussian integers.

## Proof

Let J=〈α,β〉 be the ideal generated by α and β in the ring Z[i]. Suppose δ is one of the non-zero elements of J such that N(δ) is of minimum non-zero norm and α=δγ+ψ and β=δγ 1+ψ 1 as in Theorem 11.6.2. We claim that ψ and ψ 1 are both zero. If not, then the remainders ψ and ψ 1 are in J with N(ψ)<N(δ) and N(ψ 1)<N(δ), which implies a contradiction. Hence ψ=0=ψ 1 shows that α=δγ and β=δγ 1. Consequently, δ is a common divisor of α and β. Again, since δJ, it has the form δ=λα+μβ, λ,μZ[i]. Thus δ is a multiple of every common divisor of α and β. This proves that δ is the greatest common divisor of α and β. □

The rest of the method of decomposition of Gaussian integers can be proved by the method of rational integers.

## Proposition 11.6.3

(a) If λ is a non-zero non unit Gaussian integer, then there exists a Gaussian prime p such that p divides λ in Z[i].

(b) If λ is Gaussian prime and α, β are Gaussian integers such that λ|αβ, then λ|α or λ|β in Z[i].

## Proof

Left as an exercise. □

## Theorem 11.6.3

Every non-zero, non unit Gaussian integer α can be expressed uniquely as a product α=λ 1 λ 2λ n of prime Gaussian integers, such that any other decomposition of α in Z[i] into prime Gaussian integers has the same number of factors and can be arranged in such a way that the corresponding placed factors are associates.

## Proof

Left as an exercise. □

## Theorem 11.6.4

A complex number α in the field Q(i) is a Gaussian integer iff the monic irreducible polynomial equation satisfied by α over Q has integral coefficients.

## Proof

Let α=a+bi be a Gaussian integer which is not a rational integer. Then b≠0 and α satisfies a monic irreducible quadratic equation: x 2−2ax+(a 2+b 2)=0 over Q with rational integers as coefficients. Conversely, let a number α=p+qiQ(i) satisfy a monic irreducible polynomial f(x) in Q[x] with integral coefficients. Then it follows that α is a Gaussian integer. □

## 11.7 Algebraic Number Fields

An algebraic number field is a field which is obtained from the field Q by adjoining a finite number of algebraic numbers. In this section we shall give some basic properties of algebraic number fields.

## Definition 11.7.1

An algebraic number field K is a subfield of C of the form K=Q(α 1,α 2,…,α n ), where α 1,α 2,…,α n are algebraic numbers.

## Example 11.7.1

(i) $$\mathbf{Q}(\sqrt{2},\sqrt{7},\sqrt{11},\sqrt{13})$$ is an algebraic number field.

(ii) $$\mathbf{Q}(\sqrt[17]{5},\frac{1}{\sqrt{5}})$$ is an algebraic number field.

(iii) Q(π 2) is not an algebraic number field.

## Proposition 11.7.1

The roots of unity which lie in any given algebraic number field form a cyclic group.

## Proof

Left as an exercise. □

## Remark

For more results see Exercises-IV.

Explicit computation with algebraic integers is difficult. Instead of developing a general theory, we study algebraic integers in a quadratic field. More precisely, in this section we describe algebraic integers in a field Q(α), where the complex number α is root of an irreducible quadratic polynomial in Q[x].

## Definition 11.8.1

A subfield F of C is called a quadratic field iff there exists a complex number α such that F=Q(α), where α is a root of an irreducible quadratic polynomial in Q[x].

## Remark

Q(α)={a+:a,bQ} and [Q(α):Q]=2.

## Example 11.8.1

$$\mathbf{Q}(\sqrt{7})$$ is a quadratic field. This is so because x 2−7 is irreducible in Q[x].

## Theorem 11.8.1

Let F be a quadratic field. Then there exists a unique square free integer n such that $$F=\mathbf{Q}(\sqrt{n})$$.

## Proof

Let F=Q(α), where α is a root of the irreducible polynomial x 2+bx+c over Q. If α 1 and α 2 are the two values of α, then $$\alpha_{1}=\frac{-b+\sqrt{b^{2}-4c}}{2}$$ and $$\alpha_{2}=\frac{-b-\sqrt{b^{2}-4c}}{2}$$. Hence $$\alpha_{1}+\alpha_{2}=-b\in\mathbf{Q}\Rightarrow\mathbf{Q}(\alpha_{1})= \mathbf{Q}(\alpha_{2})\Rightarrow F=\mathbf{Q}(\alpha)=\mathbf{Q}(\alpha _{1})=\mathbf{Q}(\frac{-b+\sqrt{b^{2}-4c}}{2})=\mathbf{Q}(\sqrt{a})$$, where a=b 2−4cQ. But this a is not the square of a rational number, since x 2+bx+c is irreducible in Q[x] by hypothesis. Suppose a=p/q, where p,q are integers, q>0 and gcd(p,q)=1. Let l 2 be the largest square dividing pq. Then pq=l 2 n, where n (≠1) is a square free integer and $$F=\mathbf{Q}(\sqrt{a})= \mathbf{Q}(\sqrt {p/q})=\mathbf{Q}(\sqrt{pq})=\mathbf{Q}(l\sqrt{n})=\mathbf{Q}(\sqrt{n})$$. To show the uniqueness of n, let m be another square free integer such that $$F =\mathbf{Q}(\sqrt{m})$$. Then $$\mathbf{Q}(\sqrt{n})=\mathbf{Q}(\sqrt{m})\Rightarrow\sqrt{n}=x+y\sqrt{m}$$ for some $$x,y\in \mathbf{Q}\Rightarrow n=x^{2}+my^{2}+2xy\sqrt{m}$$. If xy≠0, then $$\sqrt{m}=(n-x^{2}-my^{2})/2xy$$. This is a contradiction, since $$\sqrt{m}\notin\mathbf{Q}$$, as m is square free. Hence xy=0. Again if y=0, then $$x=\sqrt{n}$$. This is again a contradiction, since $$\sqrt{n}\notin\mathbf{Q}$$ as n is square free. Consequently, x=0 and n=my 2. Again n is square free and hence y 2=1. Consequently, m=n. □

We now describe the form of the algebraic integers α in the quadratic field $$F= \mathbf{Q}(\sqrt{n})$$, where n is a square free integer.

Let $$\alpha=\frac{l+m\sqrt{n}}{d}$$, where l,m,n are integers such that d>0 and gcd(l,m,d)=1. If m=0, then α=l/dα is a quadratic integer iff d=1. Next suppose m≠0. Now α is a root of a polynomial $$x^{2}+bx+c\in\mathbf{Z}[x]\Rightarrow (\frac{l+m\sqrt{n}}{d})^{2} +b(\frac{l+m\sqrt{n}}{d})+c=0\Rightarrow l^{2}+m^{2}n+bld+cd^{2}=0$$ and 2l+bd=0, since m≠0. Let gcd(l,d)=t>1. Then there exists a prime factor p of t and hence p|l and p|dp 2|m 2 np 2|m 2, since n is square free ⇒p|m. This is a contradiction, since gcd(l,m,d)=1 and hence t=1. Thus gcd(l,d)=1⇒d|2, since 2l+bd=0⇒d=1 or d=2. If d=1, then $$\alpha=l+m\sqrt{n}$$ is a quadratic integer. If d=2, then $$\alpha=\frac{l+m\sqrt{n}}{2}\Rightarrow (\alpha-(l/2))^{2}=\frac{m^{2}}{4}n \Rightarrow\alpha$$ is a root of the quadratic equation $$x^{2}-lx+\frac{l^{2}-m^{2}n}{4}=0$$ over $$\mathbf{Z}\Rightarrow l^{2} \equiv m^{2}n\ (\operatorname{mod}4)$$, since the above coefficients are all integers. Again gcd(l,d)=1⇒gcd(l,2)=1⇒l is odd $$\Rightarrow1\equiv m^{2}n\ (\operatorname{mod}4)$$, since $$l^{2}=m^{2}n\ (\operatorname{mod}4)$$. We now consider the possible cases: $$n\equiv1\ (\operatorname{mod}4)$$ or $$n\equiv2\ (\operatorname{mod}4)$$ or $$n\equiv3\ (\operatorname{mod}4)$$.
Case 1

If $$n\equiv1\ (\operatorname{mod}4)$$, then $$1\equiv m^{2}\ (\operatorname{mod}4)$$. This is true iff m is odd.

Case 2

If $$n\equiv2\ (\operatorname{mod}4)$$, then $$1\equiv 2m^{2}\ (\operatorname{mod}4)$$. But there is no integer m such that $$1\equiv2m^{2}\ (\operatorname{mod}4)$$.

Case 3

If $$n\equiv3\ (\operatorname{mod}4)$$, then $$1\equiv 3m^{2}\ (\operatorname{mod}4)$$. But there is no integer m such that $$1\equiv 3m^{2}\ (\operatorname{mod}4)$$.

From the above discussion there follows Theorem 11.8.2.

## Theorem 11.8.2

Let $$\mathcal{A}$$ be the set of all algebraic integers in the quadratic field $$F=\mathbf{Q}(\sqrt{n})=\{a+b\sqrt{n}: a,b\in \mathbf{Q}\}$$, where n is a square free integer. Then the set $$\mathcal{A}$$ is given by
$$\mathcal{A}=\left \{ \begin{array}{l@{\quad}l} l+m\sqrt{n}, & \textit{where }l,m \textit{ are integers and } n\not \equiv1\ (\operatorname{mod}4)\\ \frac{l+m\sqrt{n}}{2}, & \textit{where } l,m \textit{ are odd integers and } n\equiv1\ (\operatorname{mod}4). \end{array} \right .$$

## Exercises-IV

1. 1.

Show that a simple transcendental extension of a field F is isomorphic to the field of rational functions over F and is of infinite degree over F; moreover, any two simple transcendental extensions of F are isomorphic.

2. 2.
Let F(α) be a simple algebraic extension of a field F. Prove that
1. (a)

F(α)≅F[x]/〈f(x)〉, where f(x) is the minimal polynomial of α over F;

2. (b)

If n is the degree of f(x), then S={1,α,α 2,…,α n−1} forms a basis of F(α) over F;

3. (c)

If β is an algebraic element of F(α), then the degree of β over F is a divisor of degree of α.

3. 3.
1. (a)

Let F be a subfield of C. If αC and βC are algebraic over F, show that there exists an element γC such that F(α,β)=F(γ).

2. (b)

Let F be a subfield of C. If α 1,α 2,…,α n C are algebraic over F, show that there exists an element γC such that F(α 1,α 2,…,α n )=F(γ).

3. (c)

If F is an algebraic number field, prove that there exists an algebraic integer γ such that F=Q(γ).

[Hint. Use 3(b).]

4. (d)

Let F=Q(γ) be an algebraic number field, where γ is an algebraic integer. If the degree of the minimal polynomial of γ over Q is n, then prove that every element of F can be expressed uniquely in the form a 0+a 1 γ+⋯+a n−1 γ n−1, where a i are rational numbers.

[Hint. F is vector space over Q of dimension n.]

5. (e)

Let F be an algebraic number field and K be the set of all algebraic integers. Then show that the set H=KF is an integral domain (called the ring of integers of the algebraic number field K) and the quotient field of H is F.

6. (f)
Let F be a field of characteristic 0.
1. (i)

If α,β are algebraic over F, show that there exists an element γF(α,β) such that F(α,β)=F(γ).

2. (ii)

Prove that any finite field extension of the field F is a simple extension.

[Hint. Use induction argument to case (a).]

4. 4.

Let F be field and K be a field extension of F of prime dimension p. If α is an element of K but not in F, show that α has degree p and K=F(α).

[Hint. [K:F]=p⇒[K:F(α)][F(α):F]=p⇒[K:F(α)]=1, since [F(α):F]≠1, as αF. Hence [F(α):F]=p and K=F(α).]

5. 5.
Let F be a subfield of the field C. If an element α of C is algebraic over F, then prove that
1. (a)

every element β of F(α) is algebraic over F;

2. (b)

degree of β over F≤ degree of α over F.

6. 6.

Let FLK be a tower of fields such that K is algebraic over L and L is algebraic over F. Prove that K is algebraic over F.

7. 7.

Let F be field. Prove that F is algebraically closed iff every irreducible polynomial in F[x] is of degree 1.

8. 8.
1. (a)

Let F be field. Show that there exists an algebraic field extension K of F such that K is algebraically closed (called an algebraic closure of F).

2. (b)

If K and L are two algebraic closures of F, then show that there is an isomorphism ψ:KL such that ψ(a)=a, ∀aF.

Remark  Given a field F, its algebraic closure is unique up to isomorphisms.

9. 9.
1. (a)

Prove that a number α is an algebraic integer iff the additive group generated by all the powers 1,α,α 2,α 3,… of α can be generated by a finite number of elements.

2. (b)

If all the positive powers of an algebraic number α lie in an additive group generated by a finite set of numbers α 1,α 2,…,α n , then show that α is an algebraic integer.

3. (c)

Prove that the set of all algebraic integers is an integral domain.

4. (d)

In any field of algebraic numbers, prove that the set of algebraic integers is also an integral domain.

10. 10.

Show that the minimal polynomial of an algebraic integer is monic with integral coefficients.

11. 11.
1. (a)

Let K be finite extension over F such that [K:F]=n. Prove that every element α of K has a degree m over F such that m is a divisor of n.

2. (b)

Let f(x) be an irreducible cubic polynomial over a field F. If K is an extension of F of degree 2 n , then show that f(x) is irreducible in K[x].

3. (c)
Using (b) as the algebraic basis show that by straight edge and compass alone, it is impossible to
1. (i)

duplicate a general cube (duplication of a cube);

2. (ii)

trisect a general angle (trisection of an angle).

If a real number can be constructed by straight edge and compass alone, then the number is said to be constructible.

[Hint. (a) The element α generates a simple extension F(α) of F. Hence [K:F(α)][F(α):F]=n.

(b) If f(x) is a reducible polynomial over K of degree 2 n , then the cubic polynomial f(x) must have at least one linear factor in K[x]. This shows that K contains a root α of f(x). But such an element α of degree 3 over F cannot lie in a field K of degree 2 n over F by (a).

(c) (i) If it possible to construct another cube of double the volume of a unit cube, then the side x of the new cube must satisfy the equation x 3−2=0. But by the Eisenstein criterion, the polynomial x 3−2 is irreducible in Q[x]. Over any field F corresponding to the straight edge and compass construction, the polynomial x 3−2 is also irreducible by (b).

(ii) A 60 angle is constructible, because cos60 is half. But if an angle of 60 is trisected by straight edge and compass, then cos60=4cos320−3cos20 shows that x=cos20 must satisfy the cubic equation 8x 3−6x−1=0 in Q[x]. But 8x 3−6x−1 is irreducible in Q[x].]

12. 12.
An extension K of a field F is a root field of a polynomial f(x) of degree ≥1, with coefficients in F iff
1. (i)

f(x) can be factored into linear factors f(x)=a(xα 1)⋯(xα n ) in K[x]; and

2. (ii)

K=F(α 1,α 2,…,α n ).

Prove that
1. (a)

any polynomial over any field has a root field (existence).

2. (b)

all root fields of a given polynomial over a given field F are isomorphic (uniqueness).

3. (c)

for any prime integer p and any positive integer n, there exists a finite field with q=p n elements, which is the root field of x q x over Z p .

## Exercises-V

Identify the correct alternative(s) (there may be more than one) from the following list:
1. 1.
Let R be the quotient ring Z[i]/n Z[i]. Then R is an integral domain if n is
\begin{aligned} &\text{(a)}\ 19\qquad \text{(b)}\ 7\qquad \text{(c)}\ 13\qquad \text{(d)}\ 1. \end{aligned}

2. 2.
The field extension $$\mathbf{Q}(\sqrt{2}+\sqrt[3]{2})$$ over the field $$\mathbf{Q}(\sqrt{2})$$ has degree
\begin{aligned} &\text{(a)} \ 1\qquad \text{(b)} \ 3\qquad \text{(c)} \ 2\qquad \text{(d)} \ 6. \end{aligned}

3. 3.
If the polynomial x 4+x+6 has a root of multiplicity greater than 1 over a field of characteristic p (prime), then p is
\begin{aligned} &\text{(a)} \ p=5\qquad \text{(b)} \ p=3\qquad \text{(c)} \ p=2\qquad \text{(d)} \ p=7. \end{aligned}

4. 4.
Let F be a field of eight elements and B be a subset of F such that B={xF:x 7=1 and x n ≠1 for all positive integers n less than 7}. Then the number of elements in B is
\begin{aligned} &\text{(a)} \ 3\qquad \text{(b)} \ 2\qquad \text{(c)} \ 1\qquad \text{(d)} \ 6. \end{aligned}

5. 5.
Let f(x)=2x 2+x+2 and g(x)=x 3+2x 2+1 be two polynomials over the field Z 3. Then
1. (a)

f(x) and g(x) are both irreducible;

2. (b)

neither f(x) nor g(x) is irreducible;

3. (c)

f(x) is irreducible, but g(x) is not;

4. (d)

g(x) is irreducible, but f(x) is not.

6. 6.
Let F be a field and the polynomial f(x)=x 3−312312x+123123 be irreducible in F[x]. Then
1. (a)

F is a finite field with 7 elements;

2. (b)

F is a finite field with 13 elements;

3. (c)

F is a finite field with 3 elements;

4. (d)

F=Q of rational numbers.

7. 7.
Let ω be an imaginary cube root of unity i.e., ω is a complex number such that ω 3=1 and ω≠1. If K is the field $$\mathbf{Q}(\sqrt[3]{2},\omega)$$ generated by $$\sqrt[3]{2}$$ and ω over the field Q of rational numbers and n is the number of subfields L of K such that Q⊆̷L⊆̷K, then n is
\begin{aligned} &\text{(a)} \ 4\qquad \text{(b)} \ 3\qquad \text{(c)} \ 5\qquad \text{(d)} \ 2. \end{aligned}

8. 8.
Let $$F_{5^{n}}$$ be the finite field with 5 n elements. If $$F_{5^{n}}$$ contains a non-trivial root of unity, then n is
\begin{aligned} &\text{(a)} \ 15\qquad \text{(b)} \ 6\qquad \text{(c)} \ 92\qquad \text{(d)} \ 30. \end{aligned}

9. 9.
Which of the following statements is (are) correct?
1. (a)

sin7 is algebraic over Q;

2. (b)

sin−11 is algebraic over Q;

3. (c)

cos(π/17) is algebraic over Q;

4. (d)

$$\sqrt{2}+\sqrt{\pi}$$ is algebraic over Q(π).

10. 10.
Which of the following statements is (are) correct?
1. (a)

There exists a finite field in which the additive group is not cyclic;

2. (b)

Every infinite cyclic group is isomorphic to the additive group of integers;

3. (c)

If F is a finite field, there exists a polynomial p over F such that p(x)=0 for all xF, where 0 denotes the zero in F;

4. (d)

Every finite field is isomorphic to a subfield of the field of complex numbers.

11. 11.
Consider the ring Z n for n≥2. Which of the following statements is (are) correct?
1. (a)

If Z n is a field, then n is a composite integer;

2. (b)

If Z n is a field iff n is a prime integer;

3. (c)

If Z n is an integral domain, then n is prime integer;

4. (d)

If there is an injective ring homomorphism of Z 5 to Z n , then n is prime.

12. 12.
Let F p be the field Z p , where p is a prime. Let F p [x] be the associated polynomial ring. Which of the following quotient rings is (are) fields?
\begin{aligned} &\text{(a)} \quad F_5 [x]/\langle x^2 + x + 1\rangle;\qquad \text{(b)} \quad F_2 [x]/\langle x^3 + x + 1\rangle; \\ &\text{(c)} \quad F_3 [x]/\langle x^3 + x + 1\rangle;\qquad \text{(d)} \quad F_7 [x]/\langle x^2 + 1\rangle. \end{aligned}

We refer the reader to the books (Adhikari and Adhikari 2003, 2004; Alaca and Williams 2004; Artin 1991; Birkhoff and Mac Lane 2003; Hardy and Wright 2008; Hungerford 1974; Rotman 1988) for further details.

## References

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