The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

[IV, Proposition 6.3.1, EGA]

Lemma 10.157.1. We have

\[ \text{depth}(M \otimes _ R N) = \text{depth}(M) + \text{depth}(N/\mathfrak m_ RN) \]

where $R \to S$ is a local homomorphism of local Noetherian rings, $M$ is a finite $R$-module, and $N$ is a finite $S$-module flat over $R$.

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 10.71.7. According to Lemma 10.98.1 the element $f \in S$ is a nonzerodivisor on $N$ and $N/fN$ is flat over $R$. Hence by induction on $n$ we have

\[ \text{depth}(M \otimes _ R N/fN) = \text{depth}(M) + \text{depth}(N/(f, \mathfrak m_ R)N). \]

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

\[ 0 \to M \otimes _ R N \to M \otimes _ R N \to M \otimes _ R N/fN \to 0 \]

is exact where the first map is multiplication by $f$ (Lemma 10.38.12). Hence by Lemma 10.71.7 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

\[ \text{depth}(M/fM \otimes _ R N) = \text{depth}(M/fM) + \text{depth}(N/\mathfrak m_ RN). \]

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 10.71.7 and we conclude that equality holds in the formula of the lemma. $\square$


Comments (2)

Comment #3241 by Dario WeiƟmann on

Typo in the first sentence of the proof: ...and the depth of


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