This account summarises (Rosen 1995).

Abel proceeded in two steps: first he showed that a field *F* that contains all the roots of a polynomial and is contained in a radical tower can be the top of a radical tower—this is the gap in Ruffini’s analysis. Second, he showed that for polynomials of degree 5 no such field *F* is the top of a radical tower.

If

*m* = 5 then we can write

$$\displaystyle \begin{aligned}R^{1/5} = v = r_0 + r_1 x + r_2 x ^2 + \cdots + r_4 x ^4,\end{aligned}$$

where the

*r*s are symmetric functions in the roots

*x*_{1}, …,

*x*_{5} and

$$\displaystyle \begin{aligned}x = s_0 + s_1 R^{1/5} + s_2 R^{2/5} + s_3 R^{3/5} + s_4 R^{4/5}.\end{aligned}$$

As before, one deduces that

$$\displaystyle \begin{aligned}s_1 r^{1/5} = \frac{1}{5}(x_1 + \alpha^4 x_2 + \alpha^3 x_3 + \alpha^2 x_4 + \alpha x_5).\end{aligned}$$

But this is impossible, because

*s*_{1}*R*^{1∕5} takes only 5 values as the roots are permuted, but the expression on the right-hand side takes 120 values.

^{11}The case

*m* = 2 is a little longer to treat. We now have

$$\displaystyle \begin{aligned}R^{1/2} = p + qs,\; \mathrm{where} \; s= \prod_{i<j} (y_i - y_j)\end{aligned}$$

and

*p* and

*q* are symmetric functions of the roots. Indeed, we also have

\(-\sqrt {R} = p - qs\), so by simple algebra,

*p* = 0 and

\(\sqrt {R} = qs\). We may write any expression of the form

$$\displaystyle \begin{aligned}\left(p_0 + p_1 R^{\frac{1}{2}} + p_2 R^{\frac{1}{2}} + \cdots \right)^{\frac{1}{m}}\end{aligned}$$

—anything we get by adjoining

\(R^{\frac {1}{2}}\) and then passing to the next adjunction—in the form

$$\displaystyle \begin{aligned}\left(\alpha + \beta \sqrt{s^2}\right)^{\frac{1}{m}},\end{aligned}$$

where

*α* and

*β* are symmetric functions. So the solutions will be algebraic expressions in this square root. Suppose

$$\displaystyle \begin{aligned} r_1 = \left(\alpha + \beta (s^2)^{\frac{1}{2}}\right)^{\frac{1}{m}} \; \mathrm{and} \; r_2 = \left(\alpha - \beta (s^2)^{\frac{1}{2}}\right)^{\frac{1}{m}}.\end{aligned} $$

Abel noted that (

*α*^{2} −

*β*^{2}*s*^{2}) is symmetric and claimed that

*rr*_{1} = (

*α*^{2} −

*β*^{2}*s*^{2})

^{1∕m} must be symmetric, because if it not then it must be the case that

*m* = 2, but then

*r*_{1}*r*_{2} has 4 values, which is impossible. But if

*rr*_{1} is symmetric, say

*rr*_{1} =

*v*, then

$$\displaystyle \begin{aligned}r_1 + r_2 = \left(\alpha + \beta s^{\frac{1}{2}}\right)^{\frac{1}{m}} + \gamma \left(\alpha - \beta s^{\frac{1}{2}}\right)^{\frac{-1}{m}} = p_0 = R^{1/m} + \gamma R^{-1/m}.\end{aligned} $$

In this expression, replace

\(\mathbb {R}^{1/m}\) by

\(\alpha ^j \mathbb {R}^{1/m}, \; 1 \leq j \leq m-1\), where

*α* is a primitive

*m*th root of unity, and let the value of the expression corresponding to

*α*^{j} be denoted

*p*_{j}. Then

$$\displaystyle \begin{aligned}(p - p_0)(p - p_1) \ldots (p - p_{m-1}) = p^m - A_0 p^{m-1} + A_2 p^{-2} + \cdots .\end{aligned}$$

Then coefficients

*A*_{j} are rational functions in the coefficients and therefore symmetric functions in the roots. To Abel , at least, the equation was “evidently” irreducible. Therefore there are

*m* values for the

*p*_{j} and therefore also

*m* = 5. It follows by the now-familiar argument, on noting that

$$\displaystyle \begin{aligned}R^{\frac{1}{2}} = \frac{1}{5} \left(y_1 + \alpha ^4 y_2 + \alpha^3 y_3 + \alpha^2 y_4 + \alpha y_5 \right) = \left(p+ p_1 S^{\frac{1}{2}} \right)^{\frac{1}{5}}, \end{aligned}$$

that an expression of the form

*t*_{1}*R*^{1∕5} on the one hand takes 10 values (5 for the five roots, 2 for the square root) and on the other hand takes 120 values. This contradiction concludes the paper.

It seems to me that Abel did not argue that the best-possible resolvent is not up to the task (say, by showing that it only solves a restricted class of quintic equations), rather, he showed that no suitable resolvent exists, and he did so via the theory of permutations.

I conclude with some remarks about the principal obscurity in Ruffini’s work, which Abel dealt with. The element

\(\alpha = 2 \cos (2 \pi /7)\) satisfies the polynomial equation

$$\displaystyle \begin{aligned}x^3 + x^2 - 2x -1 = 0,\end{aligned}$$

which is irreducible over

\(\mathbb {Q}\), the field of rational numbers. Over the field

\(\mathbb {Q}(\alpha )\) the polynomial splits completely, but the extension

\(\mathbb {Q} : \mathbb {Q}(\alpha )\) is not a radical extension because the polynomial is not a pure cubic.

On the other hand, Abel proved that if a field

*K* contains enough roots of unity (think of it as

\(\mathbb {Q}\) with such roots of unity adjoined) and

*f*(

*x*) is a polynomial with its coefficients in the field

*K*, then the smallest field

*L* that contains all the roots of

*f*(

*x*) will arise in a sequence of radical extensions. In symbols, let

$$\displaystyle \begin{aligned}\mathbb{Q} < K < \cdots < E_j < E_{j+1} < \cdots < E_n\end{aligned}$$

be a chain of radical extensions, and suppose that

*E*_{n−1} <

*L* <

*E*_{n}. Then in fact

*L* =

*E*_{n}.

So in the presence of enough roots of unity a field contained in a chain of radical extensions is a radical extension itself, but without the appropriate roots of unity this is not necessarily the case. In particular, in the presence of enough roots of unity an extension of degree 5 must be obtained as a pure fifth root. This opened the way for Abel to concentrate on the case where the field with the roots of an irreducible quintic adjoined is precisely the top of a chain of radical extensions, and to show that this was impossible.

I omit the proofs, which will be found in (Rosen 1995) in the language of Galois theory.