Just a brief note in which I aim to decode the ideas of the proof that the Galois group of an integer polynomial reduced modulo p is a subgroup of the Galois group of the polynomial over the rationals.

Here is our idea. Let $f(x) = (x-a_1)...(x-a_n)$ be a polynomial with distinct roots in a splitting field $L/\mathbb{Q}$. Then instead of viewing the Galois group of $f$ over $\mathbb{Q}$ as something mysterious about $L$, we concretely identify it as a subgroup $G$ of the symmetric group on n letters, corresponding to its action on the roots. Note that this is a perfectly sensible thing to do, since any element of the Galois group which fixes all the roots of $f$ acts as the identity on $L$, by the definition of a splitting field.

It helps us to visualise this algebraically, so let $F(X,T_1,...,T_n) = \prod_{\sigma \in S_n} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n))$, a `generic resolvent’ which contains all the possible permutations and is therefore stable under any permutation. In particular, it is stable under the subgroup $G$ and hence has rational coefficients. Furthermore, if we factorise it $F=F_1 F_2... F_k$ into irreducibles (over $\mathbb{Q}$) these correspond to cosets of $G$.

Let us see why. On the one hand, if $(X - (a_1 T_1 + ... + a_n T_n))| F_1$ then since $G$ stabilises $F_1$, $(X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n))| F_1$ for all $\sigma \in G$. On the other hand, $\prod_{\sigma \in G} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n)$ is a rational polynomial, by Galois theory, so since $F_1$ is irreducible, we have in fact

$F_1 = \prod_{\sigma \in G} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n)$.

Similarly, the other irreducible factors represent G-cosets by exactly the same arguments. Therefore we can literally read off the Galois group by looking at the factorisation of a single irreducible factor $F_1$. What a pleasant way to study the Galois group of a polynomial!

We can use a couple of algebraic tricks to push this idea further over the field $\mathbb{Q}$ in the case where $f$ is a monic integer polynomial. Firstly, by Gauss’ Lemma we can take $F_1,F_2,...,F_k$ to be integer polynomials (pass from base ring $\mathbb{Q}(T_1,...,T_n)$ to $\mathbb{Z}[T_1,...,T_n]$). Next, now we have integer coefficients, we can reduce the whole setup modulo a prime $latex p$, and we are now studying the Galois group of $f$ over the field $\mathbb{F}_p$, but all our above arguments about irreducible factors correspond to our reading off subgroups still work.

However, note that since $F_1$ is a $\mathbb{Z}$-polynomial, when we reduce the whole game modulo p, it becomes a perfectly respectable $\mathbb{F}_p$-polynomial, and therefore we deduce that the irreducible factor of $F$ reduced modulo $p$ which is divisible in a splitting field over $\mathbb{F}_p$ by $(X - (a_1 T_1 + ... + a_n T_n))$ is an irreducible factor of the reduction of $F_1$, and therefore has factors corresponding to a subset of the factors of $F_1$. Thus, reading off the Galois group from the irreducible factors now, we deduce that

$\text{Gal}_{\mathbb{F}_p}(f ) \hookrightarrow \text{Gal}_{\mathbb{Q}}(f ).$

I think this reading off the Galois group by looking (in a splitting field) at the linear factors of the irreducibles can make one feel a little suspicious, but such suspicions are completely unfounded: by identifying the Galois group with the relevent subset of $S_n$ (having fixed an ordering of the roots in a splitting field), it is completely rigorous and elegantly transparent.