Finding Galois extension whose Galois group is Sn

Question

I want to prove that given any integer n, we can find a finite Galois extension K over $\mathbb Q$ such Gal$({K}: {\mathbb Q })$ = $S_n$

For prime p, I know finding a polynomial with exactly 2 nonreal roots will have Galois group of splitting field $S_p$.

Can we find infinitely many such polynomials for prime p?

What about composite n?

Answer 1

One alternative approach (based on Dedekind's theorem, see also here for a local explanation) is the following:

1. Select two distinct prime numbers $q$ and $r$.
2. Select an irreducible monic polynomial $g_1(x)\in\Bbb{Z}_q[x]$ of degree $n$.
3. Select an irreducible monic polynomial $g_2(x)\in\Bbb{Z}_r[x]$ of degree $n-1$.
4. Solve the instance of Chinese Remainder Theorem and find a monic polynomial $\tilde{g}(x)\in\Bbb{Z}_{qr}[x]$ such that $$\tilde{g}(x)\equiv g_1(x)\pmod q$$ as well as $$\tilde{g}(x)\equiv x g_2(x)\pmod r.$$
5. Find a "lift" $g(x)\in\Bbb{Z}[x]$ such that $g(x)$ is monic, congruent to $\tilde{g}(x)$ modulo $qr$ and has exactly two non-real zeros.

With this in place the splitting field of $g(x)$ will have Galois group $G=S_n$ over $\Bbb{Q}$:

• By item 2 $G$ contains an $n$-cycle. In particular $G$ is transitive and $g(x)$ is irreducible over $\Bbb{Q}$.
• By item 3 $G$ also contains an $(n-1)$-cycle. Together with the previous bullet this implies that $G$ is doubly transitive.
• By item 5 $G$ contains a $2$-cycle. By the previous bullet $G$ contains all the $2$-cycles, and hence $G=S_n$.

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• If item 5 is not constructive enough, then you can select a third prime $\ell>n$. Select an irreducible monic quadratic $g_3(x)\in\Bbb{Z}_\ell[x]$ and, instead of 5, add the requirement that $$ g(x)\equiv g_3(x)\prod_{i=0}^{n-3}(x-i)\pmod\ell. $$ Dedekind's theorem again implies that $G$ contains a $2$-cycle.
• See this nice post for an approach to item 5.