Lemma 15.43.1. Let $A$ be a Noetherian local ring. Then $\dim (A) = \dim (A^\wedge )$.
Proof. By Algebra, Lemma 10.97.4 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 10.52.12 this implies that
\[ \text{length}_ A(A/\mathfrak m^ n) = \text{length}_{A^\wedge }(A^\wedge /(\mathfrak m^\wedge )^ n) \]
for all $n \geq 1$. Thus $d(A) = d(A^\wedge )$ and we conclude by Algebra, Proposition 10.60.9. An alternative proof is to use Algebra, Lemma 10.112.7. $\square$
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