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Let $n \geq 5$. I want to show that $\exists p\in \mathbb{Q}[X]$ of degree $n$ with roots which are not possible to find via radicals and rational functions from the coefficients of $p$.

I've got the following theorem (Abel-Ruffini) written down in my notes.

(1) $\exists$ transcendental $\alpha_1,\dots\alpha_n\in\mathbb{C}$ s.t. $F=\mathbb{Q}(\alpha_1,\dots,\alpha_n)\cong \mathbb{Q}(X_1,\dots,X_n)$
(2) $S_n$ acts on $F$ by permuting the $\alpha_i$ and $F/F^{S_n}$ Galois with Galois group $S_n$

Now $F^{S_n}$ is the subfield of $F$ consisting of all the elements written as a symmetric functions in the $\alpha_i$. So if I write $p(X)=(X-\alpha_1)\dots(X-\alpha_n)$ then $p\in F^{S_n}(X)$ and $\textrm{Gal}(p)=S_n$ so in particular there is no way to find the roots of $p$ via radicals and rational functions from $F^{S_n}$, since $S_n$ not soluble.

Now my notes say I should be finished, but I haven't quite proved what I want to! How do I make the step from $F^{S_n}$ to $\mathbb{Q}$ in general?

Many thanks.

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What role does $K$ play in your theorem? – Jason DeVito May 15 '12 at 14:24
The theorem says there is no general "formula" in terms of the coefficients. That does not mean that there is specific $p$ with say integer coefficients. (Though in fact that is true too.) – André Nicolas May 15 '12 at 14:31
By $K$ I clearly meant $\mathbb{Q}$, sorry! Yeah I see that this says that there is no specific formula in terms of the coefficients. But how do I prove that there exists specific $p$ with rational coefficients? @AndréNicolas : could you expand on your tantalising bracket saying 'this is true too'? – Edward Hughes May 15 '12 at 14:38
It was proved later by Galois. One has to find specific integers for which the Galois group is not solvable. Too long for a comment, is in every text. – André Nicolas May 15 '12 at 14:56
Okay I'll look it up then! Many thanks. – Edward Hughes May 15 '12 at 15:08

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