# 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 exactly is your second line saying? Are you saying that any irreducible polynomial (over $\mathbb{Q}$) of degree $p$ with exactly two non-real roots will have a galois group equal to $S_p$? Apr 26 at 7:32
• @DionelJaime Yes, that is a known trick. When $p$ is a prime, a transitive subgroup of $S_p$ that contains a 2-cycle must be all of $S_p$. Apr 26 at 7:50

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 $$G=S_n$$.

• 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.