# Why $\mathfrak{p}A_\mathfrak{p} = 0$, where $A_{\mathfrak p}$ is the localization at the kernel $\mathfrak p$ of a surjective ring homomorphism.

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 $$\mathfrak{p}$$?

$$\mathfrak{p}\mathfrak{m}^r \subseteq \mathfrak{p}^2$$ 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 $$\mathfrak{p}A_{\mathfrak p}=0$$.
• Thank you! Can you please explain why $\mathfrak{m}^r A_\mathfrak{p} = A_\mathfrak{p}$? Apr 14, 2022 at 17:48
• 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 $(\mathfrak m A_{\mathfrak p})^r=A_{\mathfrak p}$. Apr 14, 2022 at 18:06