0
$\begingroup$

Let $A$ be a commutative, Noetherian, local ring, $\mathfrak{O}$ a discrete valuation ring and $\lambda : A \rightarrow \mathfrak{O}$ be an epimorphism. Let $\mathfrak{p}=\ker(\lambda)$, and consider the conormal $A$-module $\mathfrak{p}/\mathfrak{p}^2$. I'm reading that the condition $\mathfrak{p}/\mathfrak{p}^2$ has finite length is equivalent to the natural map $A_p \rightarrow \mathfrak{O}_{(0)}$ (from the localization of $A$ at $\mathfrak{p}$ to the field of fractions of $\mathfrak{O}$) being an isomorphism.

A finitely generated $A$-module has finite length iff it is annihilated by the power of the maximal ideal $\mathfrak{m}$, because $A/\mathfrak{m}$ is the only simple $A$-module. So $\mathfrak{p}\mathfrak{m}^r \subseteq \mathfrak{p}^2$, for some $r$, but why does this imply that $\mathfrak{p}A_\mathfrak{p} = 0$, where $A_\mathfrak{p}$ is the localization at $\mathfrak{p}$?

$\endgroup$

1 Answer 1

1
$\begingroup$

$\mathfrak{p}\mathfrak{m}^r \subseteq \mathfrak{p}^2$ implies $\mathfrak{p}A_{\mathfrak p}\mathfrak{m}^rA_{\mathfrak p} \subseteq \mathfrak{p}^2A_{\mathfrak p}$. But $\mathfrak{m}^rA_{\mathfrak p}=A_{\mathfrak p}$, so we get $\mathfrak{p}A_{\mathfrak p}=(\mathfrak{p}A_{\mathfrak p})^2$. By NAK we get $\mathfrak{p}A_{\mathfrak p}=0$.

$\endgroup$
2
  • $\begingroup$ Thank you! Can you please explain why $\mathfrak{m}^r A_\mathfrak{p} = A_\mathfrak{p}$? $\endgroup$
    – JBuck
    Apr 14, 2022 at 17:48
  • 1
    $\begingroup$ I've assumed that $\mathfrak p\subsetneq\mathfrak m$ (a field is not a DVR), and then there is an element $a\in\mathfrak m\setminus\mathfrak p$. Then $a\in\mathfrak m A_{\mathfrak p}$ and it is invertible, so $\mathfrak m A_{\mathfrak p}=A_{\mathfrak p}$ and then $(\mathfrak m A_{\mathfrak p})^r=A_{\mathfrak p}$. $\endgroup$
    – user26857
    Apr 14, 2022 at 18:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .