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In part of a proof in reading that proves $\mathcal{O}_K / \mathfrak{a} $ is finite for any nonzero ideal $\mathfrak{a} $ of $\mathcal{O}_K$.

It says that since $\mathcal{O}_K$ is a finitely generated additive abelian group we have $\mathcal{O}_K \cong \mathbb{Z}^n $ for some natural $n$. (Here we mean the group homomorphism)

It later says that $m \mathcal{O}_K \cong m \mathbb{Z}^n $ (here $m$ is natural number) and so $\mathcal{O}_K /m\mathcal{O}_K \cong (\mathbb{Z} /m \mathbb{Z})^n $ which is finite.

My question is how do we conclude the final ismomorphism of groups? Can’t we also say that $m\mathbb{Z}^n \cong \mathbb{Z}^n $ and therefore $\mathcal{O}_K /m\mathcal{O}_K \cong (\mathbb{Z}/\mathbb{Z})^n ?$ But this doesn’t seem to make sense.

Quotient groups of isomorphic groups aren’t necessarily isomorphic are they?

Anonmath101
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  • Does it happen to mention that it is using the same isomorphism? – NL1992 Aug 17 '21 at 19:32
  • It just says since $\mathcal{O}_K \cong \mathbb{Z} $ and $m\mathbb{O}_K \cong \mathbb{Z} $ we have $\mathcal{O}_K / m\mathbb{Z}^n \cong (\mathbb{Z}/ m\mathbb{Z})^n $ – Anonmath101 Aug 17 '21 at 19:42

1 Answers1

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I agree that the argument as written in your post could perhaps use a bit more detail. What the author is implicitly using is that, in the isomorphism $\varphi:\mathcal{O}_K\to\mathbb{Z}^n$, we have that $\varphi(m\mathcal{O}_K)=m\mathbb{Z}^n$. The reason for this is simply that $\varphi$ is a $\mathbb{Z}$-module morphism and that $m\in\mathbb{Z}$, so we can "pull out" the $m$ and get $\varphi(m\mathcal{O}_K)=m\varphi(\mathcal{O}_K)=m\mathbb{Z}^n$. Once we have shown this, then the result follows; indeed, if $M_1,M_2$ are Abelian groups, and $\psi:M_1\to M_2$ is an isomorphism between them, then $M_1\big/N\cong M_2\big/\psi(N)$ for any subgroup $N\leqslant M_1$. (Check this!) But you are correct that, if $N'$ is a subgroup of $M_2$ with $N'\cong N$, we cannot in general conclude that $M_1\big/N\cong M_2\big/N'$. For an extreme case, let $M_1=M_2=\bigoplus_{n\in\mathbb{Z}}\mathbb{Z}$; if $N=M_1$, and $N'$ is the subgroup $\bigoplus_{n\geqslant 0}\mathbb{Z}$ of $M_2$, then we have $N\cong N'$ even though $M_1\big/N=0$ and $M_2\big/N'\cong M_2$. See here for some discussion of a related question in the context of finite but not necessarily Abelian groups.

Atticus Stonestrom
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