Lemma 15.101.5. Let $(A, \mathfrak m)$ be a Noetherian local ring with henselization $A^ h$. Let $\mathfrak q \subset A^\wedge $ be a minimal prime with $\dim (A^\wedge /\mathfrak q) = 1$. Then there exists a minimal prime $\mathfrak q^ h$ of $A^ h$ such that $\mathfrak q = \sqrt{\mathfrak q^ hA^\wedge }$.

**Proof.**
Since the completion of $A$ and $A^ h$ are the same, we may assume that $A$ is henselian (Lemma 15.45.3). We will apply Lemma 15.101.4 to $A^\wedge \to A^\wedge /J$ where $J = \mathop{\mathrm{Ker}}(A^\wedge \to (A^\wedge )_{\mathfrak q})$. Since $\dim ((A^\wedge )_\mathfrak q) = 0$ we see that $\mathfrak q^ n \subset J$ for some $n$. Hence $J/J^2$ is annihilated by $\mathfrak q^ n$. On the other hand $(J/J^2)_\mathfrak q = 0$ because $J_\mathfrak q = 0$. Hence $\mathfrak m$ is the only associated prime of $J/J^2$ and we find that a power of $\mathfrak m$ annihilates $J/J^2$. Thus the lemma applies and we find that $A^\wedge /J = C^\wedge $ for some finite type $A$-algebra $C$.

Then $C/\mathfrak m C = A/\mathfrak m$ because $A^\wedge /J$ has the same property. Hence $\mathfrak m_ C = \mathfrak m C$ is a maximal ideal and $A \to C$ is unramified at $\mathfrak m_ C$ (Algebra, Lemma 10.150.7). After replacing $C$ by a principal localization we may assume that $C$ is a quotient of an étale $A$-algebra $B$, see Algebra, Proposition 10.151.1. However, since the residue field extension of $A \to C_{\mathfrak m_ C}$ is trivial and $A$ is henselian, we conclude that $B = A$ again after a localization. Thus $C = A/I$ for some ideal $I \subset A$ and it follows that $J = IA^\wedge $ (because completion is exact in our situation by Algebra, Lemma 10.96.2) and $I = J \cap A$ (by flatness of $A \to A^\wedge $). Since $\mathfrak q^ n \subset J \subset \mathfrak q$ we see that $\mathfrak p = \mathfrak q \cap A$ satisfies $\mathfrak p^ n \subset I \subset \mathfrak p$. Then $\sqrt{\mathfrak p A^\wedge } = \mathfrak q$ and the proof is complete. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: