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.

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