Lemma 10.61.6. Let $R$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Then $d(M) = \dim (\text{Supp}(M))$.

Proof. Let $M_ i, \mathfrak p_ i$ be as in Lemma 10.61.1. By Lemma 10.58.10 we obtain the equality $d(M) = \max \{ d(R/\mathfrak p_ i) \}$. By Proposition 10.59.8 we have $d(R/\mathfrak p_ i) = \dim (R/\mathfrak p_ i)$. Trivially $\dim (R/\mathfrak p_ i) = \dim V(\mathfrak p_ i)$. Since all minimal primes of $\text{Supp}(M)$ occur among the $\mathfrak p_ i$ (Lemma 10.61.5) we win. $\square$

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