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: 
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!)
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 $\mathscr O/\mathfrak p\_i\cong\Bbb F\_{25}$.