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.

Here are a few off the cuff comments that may help you on your way. First under any map $\mathcal{O}_K \rightarrow \mathbb{F}_p$, the set $p\mathcal{O}_K$ is sent to zero. So look at $\mathcal{O}_K/p\mathcal{O}_K$. This is isomorphic to $$\prod \mathcal{O}_K/\mathfrak{p}_i.$$ Maybe this is a good point for me to stop.
