Let $R$ be a commutative $\mathbb{Z}_p$-algebra integral closed domain, $\mathcal{O}_n=\mathbb{Z}_p[\zeta_n]$, and $\zeta_n$ be n-th roots of unity. For the tensor product $R \otimes_{\mathbb{Z}_p}\mathcal{O}_n$, could we find a basis of $\mathcal{O}_n$ such that $R \otimes_{\mathbb{Z}_p}\mathcal{O}_n \cong R^n$ as ring isomorphism, which means the canonical multiplication on the tensor product can be expressed by multiplication for each $R$-component in $\mathbb{Z}_p$-module isomorphism $R \otimes_{\mathbb{Z}_p}\mathcal{O}_n \cong R^n$? Thanks very much.
3
-
1$\begingroup$ $R^n$ is not integral if $n>1$. Therefore $R=\mathbb{Z}_p$ is a counterexample. $\endgroup$Nandor– Nandor2025-02-08 12:55:15 +00:00Commented Feb 8 at 12:55
-
1$\begingroup$ @Nandor you mean $\mathbb{Z}_p[\zeta_n]$ is an integral domain, but $\mathbb{Z}_p^n$ not. Thanks so much $\endgroup$Rellw– Rellw2025-02-08 13:19:12 +00:00Commented Feb 8 at 13:19
-
1$\begingroup$ @Nandor It would be nice if you could turn your comment into an answer. $\endgroup$Stefan Kohl– Stefan Kohl ♦2025-02-21 13:26:35 +00:00Commented Feb 21 at 13:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
$R^n$ is not integral if $n>1$. Therefore $R=\mathbb{Z}_p$ is a counterexample.
