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 $(O_K/\langle N\rangle)\cong(\mathbb{Z}/N\mathbb{Z})^d$.

  • 2
    $\begingroup$ 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 $\langle N \rangle \simeq N \mathbb{Z}^d$. $\endgroup$ – Pavel Čoupek May 3 '15 at 19:11
  • $\begingroup$ See etreseul.wordpress.com/2015/03/07/164 for a proof using lattice/Minkowski theory. $\endgroup$ – Sameer Kailasa May 4 '15 at 2:08

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.


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