# Abel-Ruffini theorem: confusion regarding statement

1. Jun 17, 2010

### winter85

Good day,

I am reading Stewart's Galois Theory (3rd ed). I'm up to chapter 8 where he starts tackling the issue of solubility by radicals.
The author considers independent complex variables $$t_1,t_2,...,t_n$$ and forms the polynomial $$F(t) = (t-t_1)(t-t_2)...(t-t_n)$$ which he calls the general polynomial of degree n. He says on p.95:
He lets $$L = \mathbb{C}(t_1,t_2,...,t_n)$$ be the field of all rational expressions in $$t_1,...,t_n$$ and defines K to be the subfield of L fixed by permutations of the roots. He remarks that in L the polynomial F(t) factorizes completely. He next defines solubility by "Ruffini radicals" as follows (he does mention it's not a standard definition): $$F(t)=0$$ is said to be solvable by Ruffini radicals if there exists a finite tower of subfields $$K = K_0 \subset K_1 \subset ... \subset K_r = L$$ where for $$j = 0,...,r-1$$ we have:
$$K_{j+1} = K_j(\alpha_j)$$ and $$\alpha_j^{n_j} \in K_j$$ for some positive integer $$n_j \geq 2$$. The rest of the chapter is devoted to prove that no such tower exists when $$n \geq 5$$.
My question is this: it seems to me he does not answer the point raised in the paragraph I quoted above. So what is the EXACT statement of what Ruffini and Abel proved? did they only prove that the generic quintic is not soluble by radicals, ie, there is no one formula that works for all quintics? or did they prove that there are specific quintics over $$\mathbb{Q}$$ whose roots are not contained in any radical extension of $$\mathbb{Q}$$ ?

I know that the second case is true, and that it implies the first. But I want to know what Abel and Ruffini themselves proved, and whether the non solubility of the generic quintic implies that there are quintics over $$\mathbb{Q}$$ whose roots are not in a radical extension of $$\mathbb{Q}$$.

A second point, the book points out to the gap in Ruffini's proof that Abel filled by the Theorem on Natural Irrationalities. The book says:
I don't understand what is meant by that, or how the definition of "solvable by Ruffini radicals" is different from the standard definition of "solvable by radicals" which can be found in other algebra books, such as Fraleigh's and Jacobson's. If a polynomial over $$\mathbb{Q}[x]$$ is solvable by radicals, how can its splitting field NOT be a radical extension? What is exactly the gap in Ruffini's proof and how did Abel fill it?

If anyone has Stewart's book or is familiar with this issue, I appreciate any help with clarifying these two points. Thank you.

Last edited: Jun 17, 2010