Finding Galois extension whose Galois group is Sn 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?

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

*

*Select two distinct prime numbers $q$ and $r$.

*Select an irreducible monic polynomial $g_1(x)\in\Bbb{Z}_q[x]$ of degree $n$.

*Select an irreducible monic polynomial $g_2(x)\in\Bbb{Z}_r[x]$ of degree $n-1$.

*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
$$\tilde{g}(x)\equiv x g_2(x)\pmod r.$$

*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 $G=S_n$.



*

*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.

