# if $P$ is a prime ideal of $O_K$, then $O_K/P$ is finite

let $$P$$ be a non-zero prime ideal of $$O_K$$, where $$K$$ is a number field (i.e. the degree $$[K:\mathbb{Q}]$$ is finite) then $$O_K/P$$ is finite. I'm working through a proof for this claim, however there is some group theory used in the proof which I don't understand.

Choose $$\alpha\in P$$, such that $$\alpha\neq 0$$, then $$N=|Nm(\alpha)|=\alpha\Pi_{i=2}^d \phi_i(\alpha)$$ where $$\phi_i$$ are the embeddings for $$\alpha$$, letting $$\phi_1$$ be the identity map. So $$N=\alpha\beta$$, where both $$\alpha,\beta\in O_k$$. Therefore $$N\in P$$, by definition of an ideal. So $$\langle N\rangle\subseteq P$$ and $$(O_K/P)\subseteq (O_K/\langle N\rangle)$$.

I understand everything up until this point, but now $$O_K\cong \mathbb{Z}^d$$, where $$d$$ is the degree of the minimal polynomial of $$\alpha$$. But I don't understand where this result comes from. Next the proof says that $$(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$$, which is finite, hence $$(O_K/P)\subseteq (O_K/\langle N\rangle)$$ must also be finite. Since $$\langle N\rangle = NO_K$$, does this mean that $$\langle N\rangle\cong N\mathbb{Z}^d$$, and then can we jump to the conclusion that $$(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$$.

• The fact that $\mathcal{O}_K$ is as an additive group isomorphic to $\mathbb{Z}^d$ follows from the integral basis theorem.For the rest,just think of $\mathcal{O}_k$ as an additive group,i.e. a $\mathbb{Z}$-module - for example, $\langle N \rangle$ can be thought of as $N$-multiples of all possible elements ($N \in \mathbb{Z}$), so not only $\langle N \rangle \simeq N\mathbb{Z}^d,$ it "sits the same way in $\mathcal{O}_k$ as $N\mathbb{Z}^d$ in $\mathbb{Z}^d$" - more formally, the isomorphism $\mathcal{O}_k\simeq \mathbb{Z}^d$ restricts to an iso $\langle N \rangle \simeq N \mathbb{Z}^d$. – Pavel Čoupek May 3 '15 at 19:11
• See etreseul.wordpress.com/2015/03/07/164 for a proof using lattice/Minkowski theory. – Sameer Kailasa May 4 '15 at 2:08

## 2 Answers

Let $$(p)=P\cap \mathbf Z$$. As $$P$$ is prime, $$p$$ is a prime number, and we have a commutative diagram: $$\begin{matrix} \mathbf Z &\mkern-12mu\hookrightarrow\mkern-12mu&\mathcal O_K\\ \downarrow&&\downarrow\\ \mathbf Z/p\mathbf Z&\mkern-12mu\hookrightarrow\mkern-12mu &\mathcal O_K/P \end{matrix}$$ $$\mathcal O_K$$ is a free $$\mathbf Z$$-module of rank $$d=[K:\mathbf Q]$$, so $$\mathcal O_K/p\mkern1.5mu \mathcal O_K$$ is a $$\mathbf Z/p\mathbf Z$$-vector space of dimension $$d$$, and its quotient $$\mathcal O_K/P$$ is a field extension of degree $$\le d$$. This proves $$\mathcal O_K/P$$ is finite and $$\lvert\mathcal O_K/P\rvert\le p^d.$$

By the theorem for finitely generated modules over a PID we know that there exists an integral basis $x_1,\ldots ,x_n$ of $\mathcal{O}_K$ such that $a_1x_1,\ldots ,a_nx_n$ is an integral basis of the ideal $P$, for some integers $a_i$. This gives a group homomorphism $$\mathcal{O}_K/P\simeq \mathbb{Z}/a_1\mathbb{Z}\times \cdots \times \mathbb{Z}/a_n\mathbb{Z},$$ so that $\mathcal{O}_K/P$ has $\mid a_1\cdots a_n\mid$ elements, and thus is finite.