Galois group of a polynomial modulo $p$

Question

It is well known that if $f(x)$ is a polynomial over $\mathbb Z$ then for every prime $p$ (not dividing the discriminant of $f$ (thanks to KConrad)) the Galois group of that polynomial mod $p$ over $\mathbb{F}_p$ embeds into the Galois group of $f$ over $\mathbb{Q}$. Where can I find a (easy) proof of this fact?

Answer 1

1. This result of Dedekind is not true for every prime $p$, but only for primes not dividing the discriminant of $f(x)$.

2. There is no “easy” proof for someone who knows only Galois theory (the setting where the result is usually first met). You can find a proof in Jacobson’s Basic Algebra I, attributed to Tate, that aims to be self-contained at the level of just Galois theory, but when I tried to read that while first learning Galois theory I could not understand what was going on. I mention Dedekind’s theorem as Theorem 4.13 here and after its statement I indicate where “elementary” proofs can be found in Jacobson and in Cox’s book on Galois theory. I eventually learned that the right context for understanding what this theorem is all about is within algebraic number theory: the real meaning of the theorem is surjectivity of the reduction map from the decomposition group at an unramified prime to the Galois group of the residue field extension at that prime. Without algebraic number theory that context makes no sense and I believe a proof outside that context will not be worthwhile. Do you know any algebraic number theory yet?

3. In algebraic number theory books (Samuel, Marcus, Lang, Janusz, Koch) the proof of surjectivity of the map from the decomposition group at a prime to the Galois group (or just automorphism group) of the residue field extension at that prime is overly complicated, relying on decomposition fields or completions. The proof of this result by Frobenius that I found by reading his Collected Works is much simpler and still short, so I wrote up that proof for myself here since it has been forgotten.

Answer 2

I'm not a number theorist, but I did find the Tate proof enlightening because it avoids the machinery of Dedekind domains, although at heart the proof is the same as in @KConrad's linked note. To the best of my understanding, which might be faulty, the only thing it uses that is not typically seen before Galois theory is that the algebraic integers (complex numbers which are roots of monic integer polynomials) form a ring and that the only rational algebraic integers are the actual integers.

The way I understand the proof is like this. Let $f$ be a monic integer polynomial which we can assume doesn't have repeated roots and suppose that $p$ is a prime not dividing the discriminant This is essential since you need $f$ to not have multiple roots modulo $p$. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $f$ in $\mathbb C$, $L=\mathbb Q(\alpha_1,\ldots, \alpha_n)$ and let $G$ be the Galois group of $L$ over $\mathbb Q$ (i.e., of $f$). We can think of $G$ as a faithful group of permutations of $\alpha_1,\ldots,\alpha_n$.

Now let $\mathcal O=\mathbb Z[\alpha_1,\ldots, \alpha_n]$. This is the one step where Tate differs from the classical proof using Dedekind domains I believe because $\mathcal O$ may be smaller than the full ring of integers in $L$ (a number theorist can correct me if I am mistaken). Note that $\mathcal O$ is $G$-invariant because $G$ permutes $\alpha_1,\ldots, \alpha_n$.

Choose a maximal ideal $\mathfrak m$ of $\mathcal O$ containing $p\mathcal O$ (where $p$ is our prime). This exists since $\mathcal O$ consists of algebraic integers and hence $1/p\notin \mathcal O$, whence $p\mathcal O$ is a proper ideal.

Then $\mathcal O/\mathfrak m$ is a finite extension of $\mathbb F_p=\mathbb Z/p\mathbb Z$ generated by the cosets $\overline{\alpha_1},\ldots, \overline{\alpha_n}$, which are distinct by the assumption that $p$ does not divide the discriminant of $f$. (This is where that assumption is used.) Let $H$ be the Galois group of $\mathcal O/\mathfrak m$ over $\mathbb F_p$ (which is the Galois group of the reduction $\overline{f}$ of $f$ modulo $p$). Then $H$ is a faithful permutation group of $\overline{\alpha_1},\ldots, \overline{\alpha_n}$.

Since $\mathcal O$ is $G$-invariant, $G$ permutes the maximal ideals of $\mathcal O$. Let $D$ be the stabilizer of $\mathfrak m$. For Dedekind domains this is called the decomposition group so it is probably ok to call it that. Then $D$ acts on $\mathcal O/\mathfrak m$ and it acts faithfully because $\alpha_i\mapsto \overline{\alpha_i}$ is a bijection and the $D$ action on both rings is determined by its action on these finite sets. It then suffices to show that the map $D\to H$ (which we just saw is injective) is onto. This is more or less proved in greater generality in @KConrad's link.

The basic idea is we can choose by the Chinese remainder theorem $\beta\in \mathcal O$ such that $\beta$ maps to a primitive element $\overline{\beta}$ of $\mathcal O/\mathfrak m$ and $\beta\in \sigma(\mathfrak m)$ whenever $\sigma\notin D$ (i.e., $\sigma(\mathfrak m)\neq \mathfrak m$). Consider the polynomial $g(x)=\prod_{\sigma\in G}(x-\sigma(\beta))$. This is a monic polynomial what is fixed by $G$ and hence has rational coefficients but it also has coefficients in $\mathcal O$ (which consists of algbraic integers) and so it has coefficients in $\mathbb Z$. So we can consider $\overline g$, the reduction of $g$ modulo $p$.

Clearly $g(\overline{\beta})=0$ and so the minimal polynomial $m$ of $\overline{\beta}$ divides $g$. Also $\overline{g(x)} = x^k\cdot \prod_{\sigma\in D}(x-\sigma(\overline \beta))$ because if $\sigma\notin D$, then $\beta\in \sigma^{-1}(\mathfrak m)$, whence $\sigma(\beta)\in \mathfrak m$. It follows that $m$ divides $\prod_{\sigma\in D}(x-\sigma(\overline \beta))$. But if $\tau\in H$ (the Galois group of $\mathcal O/\mathfrak m$), then $\tau(\overline{\beta})$ is a root of $m$ and hence one of the $\sigma(\overline \beta)$ with $\sigma\in D$. Since $\overline{\beta}$ is a primitive element, we deduce that $\sigma=\tau$ on $\mathcal O/\mathfrak m$. This finishes the proof that $H\cong D\leq G$.

Answer 3

I just came across the following exposition of Tate's proof on Dedekind's theorem: https://www.scirp.org/journal/paperinformation.aspx?paperid=85772 Perhaps it is of interest (I had no closer look so far).