Lemma 54.13.2. Let $(A, \mathfrak m)$ be a local Noetherian ring. Let $B \subset C$ be finite $A$-algebras. Assume that (a) $B$ is a normal ring, and (b) the $\mathfrak m$-adic completion $C^\wedge $ is a normal ring. Then $B^\wedge $ is a normal ring.

**Proof.**
Consider the commutative diagram

Recall that $\mathfrak m$-adic completion on the category of finite $A$-modules is exact because it is given by tensoring with the flat $A$-algebra $A^\wedge $ (Algebra, Lemma 10.97.2). We will use Serre's criterion (Algebra, Lemma 10.157.4) to prove that the Noetherian ring $B^\wedge $ is normal. Let $\mathfrak q \subset B^\wedge $ be a prime lying over $\mathfrak p \subset B$. If $\dim (B_\mathfrak p) \geq 2$, then $\text{depth}(B_\mathfrak p) \geq 2$ and since $B_\mathfrak p \to B^\wedge _\mathfrak q$ is flat we find that $\text{depth}(B^\wedge _\mathfrak q) \geq 2$ (Algebra, Lemma 10.163.2). If $\dim (B_\mathfrak p) \leq 1$, then $B_\mathfrak p$ is either a discrete valuation ring or a field. In that case $C_\mathfrak p$ is faithfully flat over $B_\mathfrak p$ (because it is finite and torsion free). Hence $B^\wedge _\mathfrak p \to C^\wedge _\mathfrak p$ is faithfully flat and the same holds after localizing at $\mathfrak q$. As $C^\wedge $ and hence any localization is $(S_2)$ we conclude that $B^\wedge _\mathfrak p$ is $(S_2)$ by Algebra, Lemma 10.164.5. All in all we find that $(S_2)$ holds for $B^\wedge $. To prove that $B^\wedge $ is $(R_1)$ we only have to consider primes $\mathfrak q \subset B^\wedge $ with $\dim (B^\wedge _\mathfrak q) \leq 1$. Since $\dim (B^\wedge _\mathfrak q) = \dim (B_\mathfrak p) + \dim (B^\wedge _\mathfrak q/\mathfrak p B^\wedge _\mathfrak q)$ by Algebra, Lemma 10.112.6 we find that $\dim (B_\mathfrak p) \leq 1$ and we see that $B^\wedge _\mathfrak q \to C^\wedge _\mathfrak q$ is faithfully flat as before. We conclude using Algebra, Lemma 10.164.6. $\square$

## 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Your email address will not be published.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)