Suppose that we have a number field $K$ and ring of integers $\mathcal{O}_K$. Let $\mathfrak{a} \subset \mathfrak{a}'$ be two finitely generated $\mathcal{O}_K$ submodules of $K$. Then the proposition states that $d(\mathfrak{a}) = (\mathfrak{a}' : \mathfrak{a}) d(\mathfrak{a}')$ where $d(\mathfrak{a})$ is the discriminant and $(\mathfrak{a}' : \mathfrak{a})$ is the index. Some online notes use this to show a useful corollary: If $K = \mathbb{Q}(\alpha)$ for some $\alpha \in \mathcal{O}_K$, then we have $$ d(1, \alpha, \ldots, \alpha^{n - 1}) = (\mathcal{O}_K : \mathbb{Z}[\alpha])^2 \Delta_K.$$

However, I don't see how this can be applied, as $\mathbb{Z}[\alpha]$ is not an $\mathcal O_K$ submodule of $K$? I am sure I am being an idiot here. Thanks for any help!

  • 1
    $\begingroup$ You have a bad typo: the proposition you state should have $(\mathfrak a':\mathfrak a)^2$. Anyway, the notion of a discriminant does not require $\mathfrak a$ and $\mathfrak a'$ to be $\mathcal O_K$-modules in $K$. Why do you think they must be? For the math to work, they can just be free $\mathbf Z$-modules $L$ and $L'$ in $K$ of full rank $n = [K:\mathbf Q]$: if $L \subset L'$ then $d(L) = [L':L]^2d(L')$. The result you want is the special case of that where $L = \mathbf Z[\alpha]$ and $L' = \mathcal O_K$. $\endgroup$
    – KCd
    Dec 30, 2022 at 4:33
  • $\begingroup$ @KCd Thanks! I figured it out by myself a few days ago that it probably holds for $\mathbb{Z}$-modules. It’s just the way Neukirch phrased the proposition which made me think that there’s something special with $\mathcal{O}_K$ modules. I’m not really familiar with linear algebra over modules so that’s why I got stuck for quite a while. btw, I’m a huge fan of your expository papers! $\endgroup$ Dec 31, 2022 at 5:13


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