
Lemma 10.62.10. Let $R$ is a Noetherian local ring, $M$ a finite $R$-module, and $f \in \mathfrak m$ an element of the maximal ideal of $R$. Then

$\dim (\text{Supp}(M/fM)) \leq \dim (\text{Supp}(M)) \leq \dim (\text{Supp}(M/fM)) + 1$

If $f$ is not in any of the minimal primes of the support of $M$ (for example if $f$ is a nonzerodivisor on $M$), then equality holds for the right inequality.

Proof. (The parenthetical statement follows from Lemma 10.62.9.) The first inequality follows from $\text{Supp}(M/fM) \subset \text{Supp}(M)$, see Lemma 10.39.9. For the second inequality, note that $\text{Supp}(M/fM) = \text{Supp}(M) \cap V(f)$, see Lemma 10.39.9. It follows, for example by Lemma 10.61.2 and elementary properties of dimension, that it suffices to show $\dim V(\mathfrak p) \leq \dim (V(\mathfrak p) \cap V(f)) + 1$ for primes $\mathfrak p$ of $R$. This is a consequence of Lemma 10.59.12. Finally, if $f$ is not contained in any minimal prime of the support of $M$, then the chains of primes in $\text{Supp}(M/fM)$ all give rise to chains in $\text{Supp}(M)$ which are at least one step away from being maximal. $\square$

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