Factoring a prime over a splitting field and factoring a polynomial over of a finite field

Question

Let $f$ be a separable monic polynomials with integer coefficient. Let $K$ be the splitting field of $f$ over $\mathbb{Q}$, and let $p$ be a prime. What is the relationship between the statements

$f$ splits over $\mathbb{F}_p$ and

$p$ is not irreducible in $K$ (there is probably some fancy term for this, like $p$ is inert in $K$?)

There are equivalent for the Gaussian Integers, so probably also for any quadratic extension with trivial class group.

I suspect they are equivalent, and this is probably extremely well known, but I have asked a couple people with no luck.

Answer 1

Let $ K = \mathbf Q(\alpha) $ be a number field with ring of integers $ \mathcal O_K $, where $ \alpha $ is an algebraic integer with minimal polynomial $ \pi(X)$. Dedekind's factorization criterion tells us that if $ p $ is a prime which does not divide the index $ [\mathcal O_K : \mathbf Z[\alpha]] $, then the way $ \pi(X) $ splits modulo $ p $ is equivalent to the way $ (p) $ splits in $ \mathcal O_K $, in other words, if we have

$$ (p) = \prod_{i=1}^n \mathfrak p_i^{e_i} $$

in $ \mathcal O_K $ where the inertia degrees of $ \mathfrak p_i $ are $ f_i $, then we have

$$ \pi(X) \equiv \prod_{i=1}^n \pi_i(X)^{e_i} \pmod{p} $$

where the $ \pi_i $ are irreducible in $ \mathbf Z/p\mathbf Z[X] $ with $ \deg \pi_i = f_i $. This is essentially a consequence of the isomorphisms

$$ \mathbf Z[\alpha]/(p) \cong \mathbf Z[X]/(\pi(X), p) \cong \mathbf Z/p \mathbf Z[X]/(\pi(X)) $$

and the fact that you can read off the relevant data from the ring structure alone. As an example of how this fails when the criterion regarding the index is not met, consider $ K = \mathbf Q(\sqrt{5}) $. $ \mathbf Z[\sqrt{5}] $ has index $ 2 $ in $ \mathcal O_K $, and we have $ x^2 - 5 = (x+1)^2 $ in $ \mathbf Z/2\mathbf Z $, which suggests that $ 2 $ would be a totally ramified prime. However, this is not the case: $ \operatorname{disc} \mathcal O_K = 5 $, therefore $ 2 $ is not even a ramified prime in $ \mathcal O_K $. (This is a counterexample to your claim, since $ \mathbf Q(\sqrt{5}) $ is actually the splitting field of $ x^2 - 5 $ over $ \mathbf Q $. Note that $ \mathcal O_K $ is even a principal ideal domain!)

Answer 2

Let’s look at a simple example, $f(X)=X^3-2$ and its splitting field $K=\Bbb Q(\lambda,\omega)$, over $\Bbb Q$, where $\lambda=\sqrt[3]2$ and $\omega$ is a primitive cube root of unity. Take $p=5$, and look at $f\in\Bbb F_5[X]$, where it factors as $(X+2)(X^2+3X+4)$, so $f$ doesn’t split there.

Calling $\mathscr O$ the ring of algebraic integers of $K$, we can see that $5\mathscr O$ factors as $\mathfrak p_1\mathfrak p_2\mathfrak p_3$, where in each case $\mathscr O/\mathfrak p_i\cong\Bbb F_{25}$.