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 be a polynomial with distinct roots in a splitting field . Then instead of viewing the Galois group of over as something mysterious about , we concretely identify it as a subgroup 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 acts as the identity on , by the definition of a splitting field.

It helps us to visualise this algebraically, so let , a `generic resolvent’ which contains all the possible permutations and is therefore stable under any permutation. In particular, it is stable under the subgroup and hence has rational coefficients. Furthermore, if we factorise it into irreducibles (over ) these correspond to cosets of .

Let us see why. On the one hand, if then since stabilises , for all . On the other hand, is a rational polynomial, by Galois theory, so since is irreducible, we have in fact

.

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 . 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 in the case where is a monic *integer* polynomial. Firstly, by Gauss’ Lemma we can take to be integer polynomials (pass from base ring to ). 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 over the field , but all our above arguments about irreducible factors correspond to our reading off subgroups still work.

However, note that since is a -polynomial, when we reduce the whole game modulo p, it becomes a perfectly respectable -polynomial, and therefore we deduce that the irreducible factor of reduced modulo which is divisible in a splitting field over by is an irreducible factor of the reduction of , and therefore has factors corresponding to a subset of the factors of . Thus, reading off the Galois group from the irreducible factors now, we deduce that

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 (having fixed an ordering of the roots in a splitting field), it is completely rigorous and elegantly transparent.

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