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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.

vukov
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  • As you guessed, “$p$ is inert in $K$” means, if $\mathscr O$ is the ring of integers of $K$, that $p\mathscr O$ is still prime in $\mathscr O$. Can’t happen (for Galois extensions) unless $K$ is cyclic over $\Bbb Q$. – Lubin Aug 20 '16 at 02:24
  • @hardmath, it’s an abuse of language that’s common enough in casual talk. – Lubin Aug 20 '16 at 02:27
  • An example of what I mean would be $f = x^2 -2$, $p=7=(3+\sqrt{2})(3-\sqrt{2})$, and $x^2 - 2 = (x-3)(x-4)$ in $\mathbb{F}_7$. – vukov Aug 20 '16 at 02:29

2 Answers2

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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 $\mathscr O/\mathfrak p_i\cong\Bbb F_{25}$.

Lubin
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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!)

Ege Erdil
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    This is great — I should have thought of this aspect of the question. – Lubin Aug 20 '16 at 03:23
  • This is more general than it seems in the sense that we can always pick a primitive element for number fields by the primitive element theorem. For instance, the splitting of $ 5 $ in $ \mathbf Q(\lambda, \omega) $ comes down to the splitting of $ x^6+3 x^5+6 x^4+3 x^3+9 x+9 $ (the minimal polynomial of $ \lambda + \omega $) in $ \mathbf Z/5\mathbf Z[X] $, since the discriminant of this polynomial is not divisible by 5. Indeed, we have the splitting $ (x^2+2) (x^2+4 x+1) (x^2+4 x+2) $ in $ \mathbf Z/5\mathbf Z[X] $. – Ege Erdil Aug 20 '16 at 03:45
  • Wow, @Starfall,I did the same computation, just to check that I hadn’t stepped into a pothole. – Lubin Aug 20 '16 at 03:54
  • I think it is true that if $\mathbb{Z}[x]/f(x)$ is a unique factorization domain then $f$ splitting $\implies p$ is not irreducible over $\mathbb{Z}[x]/f(x)$ (at least for quadratic extensions). You can just modify Dedekind's proof of Fermat's theorem on the sum of two squares. I didn't realize that $\mathbb{Z}[x]/f(x)$ isn't always isomorphic to the ring of integers. – vukov Aug 21 '16 at 16:47
  • Uh, if $ \mathbf Z[x]/(f) $ is a domain, then $ f $ has to be irreducible in $ \mathbf Z[x] $ (and if it is monic, by extension, in $ \mathbf Q[x] $), and it isn't always isomorphic to the ring of integers... I just gave an example with $ K = \mathbf Q(\sqrt{5}) $, where $ \mathcal O_K = \mathbf Z[\phi] $ but $ \mathbf Z[x]/(x^2 - 5) \cong \mathbf Z[\sqrt{5}] $. We have a pretty good idea of what rings of integers of quadratic extensions look like, so you can just apply the above criterion to see how the splitting takes place anyway... – Ege Erdil Aug 21 '16 at 17:46