# Splitting field over $\mathbb{F}_p$

Let $$f \in \mathbb{Z}[X]$$ be a monic polynomial of degree $$d$$. Let $$E$$ be the splitting field of $$f$$ over $$\mathbb{Q}$$ and let $$R$$ be the ring of integers in $$E$$. Suppose $$p$$ is a prime not dividing the discriminant $$D_f$$, let $$\bar{f} \in \mathbb{F}_p[X]$$ be its reduction modulo $$p$$ and let $$P$$ be a prime ideal of $$R$$ containing $$p$$. It's a classic result that $$R/P$$ is a finite field of order dividing $$p^n$$.

Can one show that $$R/P$$ is the splitting field of $$\bar{f}$$ over $$\mathbb{F}_p$$?

• Do you assume the irreducibility of $*f *$ over $Q$? – nguyen quang do Dec 3 '19 at 13:59
• Not necessarily. – user391447 Dec 3 '19 at 15:14
• I quoted from the following source: www2.bc.edu/mark-reeder/Galois.pdf on p.59 – user391447 Dec 3 '19 at 15:15

The additional point is the degree $$n$$ of the splitting field $$E$$ of the polynomial $$g$$. Since we are in characteristic $$0$$ and $$E/\mathbf Q$$ is normal, $$E$$ is galois of degree dividing $$d$$. Its Galois group $$G$$ permutes transitively the prime ideals $$P$$ of $$R$$ above $$p$$. It follows classically that all the $$r$$ primes $$P$$ of $$E$$ over $$p$$ have the same indices of ramification $$e$$ and of inertia $$f$$, so that $$n=ref$$. In Galois theoretic terms, let $$I \subset D$$ resp. be the inertia and decomposition subgroups of a prime $$P$$ above $$p$$. Recall that $$s \in D$$ iff $$s(P)=P$$, and $$I$$ is the kernel of the surjective homomorphism constructed as follows: given another prime $$Q$$ above $$p$$, by the definition of $$D$$, $$x\equiv y$$ mod $$P$$ implies $$s(x)\equiv s(y)$$ mod $$Q$$ for any $$s\in D$$, which allows to define naturally the quotient map $$s\in D \to\bar s \in Gal((R/P)/\mathbf F_p)$$ (the Galois group of the extension of residue fields). Obviously $$G$$ has order $$n$$, the $$G$$-orbit of $$P$$ has cardinal $$r$$, and it is classically known that $$I$$ has order $$f$$ (the proof is not absolutely immediate, see e.g. P. Samuel's ANT, chap. 66, §2), i.e. $$R/P$$ is a finite field of cardinal $$p^f$$. Note that this means that the residual polynomial $$\bar g$$ splits in $$R/P$$ because the residual extension $$(R/P)/\mathbf F_p$$ is galois.
You suppose moreover that $$p$$ does not divide the discriminant $$D(K)$$ (your notation $$D_g$$ is somewhat unusual), which is to say $$p$$ is unramified in $$K$$, hence also in $$E$$, so the inertia index $$f$$ is $$1$$. This means that $$\bar g$$ splits in $$\mathbf F_p$$ in this case.