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I am learning Galois theory and am very close to a completed proof of the insolvability of the quintic. It remains to show that a polynomial can be constructed (over a reasonable field $F$) with Galois group $S_n$. It seems to me that this is usually proven by considering the "general polynomial" of degree $n$, which I understand the following way:

The general polynomial $p(x)$ (here, of degree 3 for clarity), has roots $\alpha, \beta, \gamma$ which are indeterminate (or equivalently algebraically independent) over $F$, and as such has the form

$x^3-(\alpha +\beta +\gamma)x^2+(\alpha \beta + \beta \gamma + \alpha \gamma)x-(\alpha \beta \gamma)$

Clearly, $p(x) \notin F[x]$ by the algebraic independence (and thus transcendence) of $\alpha, \beta, \gamma$, and rather

$p(x) \in E[x] = F(\alpha +\beta +\gamma , \alpha \beta + \beta \gamma + \alpha \gamma , \alpha \beta \gamma)[x]$

These coefficients are symmetric expressions in the roots of $p(x)$, and as such are fixed by elements of $Gal(p(x))$. It is easy enough to see $Gal(p(x)) = S_n$.

Thus, we have constructed a generic polynomial with symmetric Galois group: however, we have done so over our field $F$, we have done so over this field $E$. What is the use of this? Is my understanding of what is meant by "general polynomial" flawed?

C V Astley
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  • This shows how to avoid talk of general polynomials: https://math.stackexchange.com/questions/165675/constructing-a-galois-extension-field-with-galois-group-s-n – ancient mathematician May 11 '23 at 15:26
  • The possibly confusing fact about this construction is that it does NOT give a recipe for finding a polynomial $f(x)\in F[x]$ of degree $n$ with Galois group $S_n$ TO A GIVEN FIELD $F$. What it does is the following. We are given an integer $n$. We then construct a field $F$ and a polynomial $f(x)\in F[x]$ of degree $n$ such that the Galois group of $f(x)$ over $F$ is $S_n$.. – Jyrki Lahtonen May 12 '23 at 04:24
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    (cont'd) The general polynomial means that if we start with a given field $K$, we can use $F=K(a_1,a_2,\ldots,a_n)$ such that $$f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n$$ with $a_1,a_2,\ldots,a_n$ algebraically independent transcendentals over $K$. – Jyrki Lahtonen May 12 '23 at 04:24
  • This is in a sense the best we can do. Undoubtedly you know that over the reals the only non-trivial Galois group you can get is cyclic of order two. Similarly, over $\Bbb{F}_p$ all the finite extensions have cyclic Galois groups. To get $S_n$, $n>2$ to appear as a Galois group we absolutely need to place some constraints on the field. – Jyrki Lahtonen May 12 '23 at 04:29
  • All this underlines the need to be careful when phrasing Abel-Ruffini and interpreting its meaning. – Jyrki Lahtonen May 12 '23 at 04:33

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