Lemma 15.42.1. Let $A$ be a Noetherian local ring. Then $\dim (A) = \dim (A^\wedge )$.

## 15.42 Permanence of properties under completion

Given a Noetherian local ring $(A, \mathfrak m)$ we denote $A^\wedge $ the completion of $A$ with respect to $\mathfrak m$. We will use without further mention that $A^\wedge $ is a Noetherian complete local ring with maximal ideal $\mathfrak m^\wedge = \mathfrak m A^\wedge $ and that $A \to A^\wedge $ is faithfully flat. See Algebra, Lemmas 10.96.6, 10.96.4, and 10.96.3.

**Proof.**
By Algebra, Lemma 10.96.4 the map $A \to A^\wedge $ induces isomorphisms $A/\mathfrak m^ n = A^\wedge /(\mathfrak m^\wedge )^ n$ for $n \geq 1$. By Algebra, Lemma 10.51.12 this implies that

for all $n \geq 1$. Thus $d(A) = d(A^\wedge )$ and we conclude by Algebra, Proposition 10.59.8. An alternative proof is to use Algebra, Lemma 10.111.7. $\square$

Lemma 15.42.2. Let $A$ be a Noetherian local ring. Then $\text{depth}(A) = \text{depth}(A^\wedge )$.

**Proof.**
See Algebra, Lemma 10.157.2.
$\square$

Lemma 15.42.3. Let $A$ be a Noetherian local ring. Then $A$ is Cohen-Macaulay if and only if $A^\wedge $ is so.

**Proof.**
A local ring $A$ is Cohen-Macaulay if and only if $\dim (A) = \text{depth}(A)$. As both of these invariants are preserved under completion (Lemmas 15.42.1 and 15.42.2) the claim follows.
$\square$

Lemma 15.42.4. Let $A$ be a Noetherian local ring. Then $A$ is regular if and only if $A^\wedge $ is so.

**Proof.**
If $A^\wedge $ is regular, then $A$ is regular by Algebra, Lemma 10.109.9. Assume $A$ is regular. Let $\mathfrak m$ be the maximal ideal of $A$. Then $\dim _{\kappa (\mathfrak m)} \mathfrak m/\mathfrak m^2 = \dim (A) = \dim (A^\wedge )$ (Lemma 15.42.1). On the other hand, $\mathfrak mA^\wedge $ is the maximal ideal of $A^\wedge $ and hence $\mathfrak m_{A^\wedge }$ is generated by at most $\dim (A^\wedge )$ elements. Thus $A^\wedge $ is regular. (You can also use Algebra, Lemma 10.111.8.)
$\square$

Lemma 15.42.5. Let $A$ be a Noetherian local ring. Then $A$ is a discrete valuation ring if and only if $A^\wedge $ is so.

**Proof.**
This follows from Lemmas 15.42.1 and 15.42.4 and Algebra, Lemma 10.118.7.
$\square$

Lemma 15.42.6. Let $A$ be a Noetherian local ring.

If $A^\wedge $ is reduced, then so is $A$.

In general $A$ reduced does not imply $A^\wedge $ is reduced.

If $A$ is Nagata, then $A$ is reduced if and only if $A^\wedge $ is reduced.

**Proof.**
As $A \to A^\wedge $ is faithfully flat we have (1) by Algebra, Lemma 10.158.2. For (2) see Algebra, Example 10.118.5 (there are also examples in characteristic zero, see Algebra, Remark 10.118.6). For (3) see Algebra, Lemmas 10.156.13 and 10.156.10.
$\square$

Lemma 15.42.7. Let $A \to B$ be a local homomorphism of Noetherian local rings. Then the induced map of completions $A^\wedge \to B^\wedge $ is flat if and only if $A \to B$ is flat.

**Proof.**
Consider the commutative diagram

The vertical arrows are faithfully flat. Assume that $A^\wedge \to B^\wedge $ is flat. Then $A \to B^\wedge $ is flat. Hence $B$ is flat over $A$ by Algebra, Lemma 10.38.9.

Assume that $A \to B$ is flat. Then $A \to B^\wedge $ is flat. Hence $B^\wedge /\mathfrak m_ A^ n B^\wedge $ is flat over $A/\mathfrak m_ A^ n$ for all $n \geq 1$. Note that $\mathfrak m_ A^ n A^\wedge $ is the $n$th power of the maximal ideal $\mathfrak m_ A^\wedge $ of $A^\wedge $ and $A/\mathfrak m_ A^ n = A^\wedge /(\mathfrak m_ A^\wedge )^ n$. Thus we see that $B^\wedge $ is flat over $A^\wedge $ by applying Algebra, Lemma 10.98.11 (with $R = A^\wedge $, $I = \mathfrak m_ A^\wedge $, $S = B^\wedge $, $M = S$). $\square$

Lemma 15.42.8. Let $A \to B$ be a flat local homomorphism of Noetherian local rings such that $\mathfrak m_ A B = \mathfrak m_ B$ and $\kappa (\mathfrak m_ A) = \kappa (\mathfrak m_ B)$. Then $A \to B$ induces an isomorphism $A^\wedge \to B^\wedge $ of completions.

**Proof.**
By Algebra, Lemma 10.96.7 we see that $B^\wedge $ is the $\mathfrak m_ A$-adic completion of $B$ and that $A^\wedge \to B^\wedge $ is finite. Since $A \to B$ is flat we have $\text{Tor}_1^ A(B, \kappa (\mathfrak m_ A)) = 0$. Hence we see that $B^\wedge $ is flat over $A^\wedge $ by Lemma 15.27.5. Thus $B^\wedge $ is a free $A^\wedge $-module by Algebra, Lemma 10.77.4. Since $A^\wedge \to B^\wedge $ induces an isomorphism $\kappa (\mathfrak m_ A) = A^\wedge /\mathfrak m_ A A^\wedge \to B^\wedge /\mathfrak m_ A B^\wedge = B^\wedge /\mathfrak m_ B B^\wedge = \kappa (\mathfrak m_ B)$ by our assumptions (and Algebra, Lemma 10.95.3), we see that $B^\wedge $ is free of rank $1$. Thus $A^\wedge \to B^\wedge $ is an isomorphism.
$\square$

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