Let $p$ be a prime that decomposes completely in $O_K$. Show that $\#\text{Hom}(O_K, \mathbb{F}_p)=n$ Let $K$ be a number field of degree $n$. We say a prime $p$ decomposes completely in $O_K$ if $pO_K = \mathfrak{p}_1...\mathfrak{p}_n$ for some prime ideals $\mathfrak{p}_i$. I want to show that if $p$ is a prime that decomposes completely then $\#\text{Hom}(O_K, \mathbb{F}_p)=n$ but I don't really know how to attack this problem. I feel like I should be using that $O_K$ is a finitely generated $\mathbb{Z}$-module or that the degree of the extension is n, but I don't really know how.

Hints/roadmap: 


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*So we have a homomorphism $f$ of rings from $O_K$ to the field $\Bbb{F}_p$. Can you show that $f$ is necessarily surjective?

*The previous bullet implies that $O_K/\operatorname{ker}(f)\simeq\Bbb{F}_p$. Do you recall what we can say about an ideal $I$ of a commutative ring $R$, when the quotient ring $R/I$ is an integral domain?

*Why does the choice of $\operatorname{ker}(f)$ determine $f$ uniquely in this case? Why do different choices for the kernel give rise to different homomorphisms.

