The Stacks project

Proof. We will prove this by induction on $\text{depth}_ I(M)$.

If $\text{depth}_ I(M) = 0$, then $I$ is contained in one of the associated primes $\mathfrak p$ of $M$ (Algebra, Lemma 10.63.18). Then $\mathfrak p \in \text{Supp}(M/IM)$, hence $\dim (\text{Supp}(M/IM)) \geq \dim (R/\mathfrak p) = \dim (\text{Supp}(M))$ where equality holds by Algebra, Lemma 10.103.7. Thus the lemma holds in this case.

If $\text{depth}_ I(M) > 0$, we pick $x \in I$ which is a nonzerodivisor on $M$. Note that $(M/xM)/I(M/xM) = M/IM$. On the other hand we have $\text{depth}_ I(M/xM) = \text{depth}_ I(M) - 1$ by Lemma 47.11.3 and $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$ by Algebra, Lemma 10.63.10. Thus the result by induction hypothesis. $\square$