$\operatorname{disc}(\mathbb{Z}[\alpha]) = [\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\operatorname{disc}(\mathcal{O}_K)$.

Let $\alpha\in\mathcal{O}_K$ such that $K=\mathbb{Q}[\alpha]$. Define $\operatorname{disc}(\mathbb{Z}[\alpha]) := \operatorname{disc}(1,\alpha,\dots,\alpha^{n-1})$. Show $\operatorname{disc}(\mathbb{Z}[\alpha]) = [\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\operatorname{disc}(\mathcal{O}_K)$.

I am using this fact to solve other problems, but I tried to prove it and I can't, I also found a proof in algebraic number theory by Bosman of something I think is the same, but i'm not sure because I don't understand the proof. In the book they prove that for $A\subseteq A'$ full rank lattices $\Delta(A) = (A':A)^2\Delta A'$.

• You can write the basis elements of $\Bbb{Z}[\alpha]$ as $\Bbb{Z}$-linear combinations of the basis elements of $\mathcal{O}_K$. Let $M$ be that matrix. Then $[\mathcal{O}_K:\mathbb{Z}[\alpha]]=|\det M|$ $\leftarrow$ do you know this? When you compute the discriminants using the bilinear trace form, you therefore will replace the matrix $D$ of that form with $MDM^T$, when you switch from one basis to the other. The discriminant is essentially $\det D$, so you multiply it with $(\det M)^2$. – Jyrki Lahtonen Sep 10 '17 at 20:33
• This has probably been explained here more verbosely and with better notation. But I'm calling it a day. Hope you find something. – Jyrki Lahtonen Sep 10 '17 at 20:34
• "" do you know this?" No. I understand everything you say but why does $[O_K:\mathbb{Z}[\alpha]] = |detM|$ – NuKexZ Sep 10 '17 at 22:19
• By the stacked bases theorem we can assume that $M$ is diagonal. And in that case it is easy. – Jyrki Lahtonen Sep 11 '17 at 16:06