The Stacks project

Proof. In the statement and in the proof below, we take the depth of $M$ as an $R$-module, the depth of $M \otimes _ R N$ as an $S$-module, and the depth of $N/\mathfrak m_ RN$ as an $S/\mathfrak m_ RS$-module. Denote $n$ the right hand side. First assume that $n$ is zero. Then both $\text{depth}(M) = 0$ and $\text{depth}(N/\mathfrak m_ RN) = 0$. This means there is a $z \in M$ whose annihilator is $\mathfrak m_ R$ and a $\overline{y} \in N/\mathfrak m_ RN$ whose annihilator is $\mathfrak m_ S/\mathfrak m_ RS$. Let $y \in N$ be a lift of $\overline{y}$. Since $N$ is flat over $R$ the map $z : R/\mathfrak m_ R \to M$ produces an injective map $N/\mathfrak m_ RN \to M \otimes _ R N$. Hence the annihilator of $z \otimes y$ is $\mathfrak m_ S$. Thus $\text{depth}(M \otimes _ R N) = 0$ as well.

Assume $n > 0$. If $\text{depth}(N/\mathfrak m_ RN) > 0$, then we may choose $f \in \mathfrak m_ S$ mapping to $\overline{f} \in S/\mathfrak m_ RS$ which is a nonzerodivisor on $N/\mathfrak m_ RN$. Then $\text{depth}(N/\mathfrak m_ RN) = \text{depth}(N/(f, \mathfrak m_ R)N) + 1$ by Lemma 10.72.7. According to Lemma 10.99.1 the element $f \in S$ is a nonzerodivisor on $N$ and $N/fN$ is flat over $R$. Hence by induction on $n$ we have

Because $N/fN$ is flat over $R$ the sequence

is exact where the first map is multiplication by $f$ (Lemma 10.39.12). Hence by Lemma 10.72.7 we find that $\text{depth}(M \otimes _ R N) = \text{depth}(M \otimes _ R N/fN) + 1$ and we conclude that equality holds in the formula of the lemma.

If $n > 0$, but $\text{depth}(N/\mathfrak m_ RN) = 0$, then we can choose $f \in \mathfrak m_ R$ which is a nonzerodivisor on $M$. As $N$ is flat over $R$ it is also the case that $f$ is a nonzerodivisor on $M \otimes _ R N$. By induction on $n$ again we have

In this case $\text{depth}(M \otimes _ R N) = \text{depth}(M/fM \otimes _ R N) + 1$ and $\text{depth}(M) = \text{depth}(M/fM) + 1$ by Lemma 10.72.7 and we conclude that equality holds in the formula of the lemma. $\square$