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*}

Lemma 10.65.2. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime of $R$. The following are equivalent:

  1. $\mathfrak p$ is weakly associated to $M$,

  2. $\mathfrak pR_{\mathfrak p}$ is weakly associated to $M_{\mathfrak p}$, and

  3. $M_{\mathfrak p}$ contains an element whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$.

Proof. Assume (1). Then there exists an element $m \in M$ such that $\mathfrak p$ is minimal among the primes containing the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$. As localization is exact, the annihilator of $m$ in $M_{\mathfrak p}$ is $I_{\mathfrak p}$. Hence $\mathfrak pR_{\mathfrak p}$ is a minimal prime of $R_{\mathfrak p}$ containing the annihilator $I_{\mathfrak p}$ of $m$ in $M_{\mathfrak p}$. This implies (2) holds, and also (3) as it implies that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$.

Applying the implication (1) $\Rightarrow $ (3) to $M_{\mathfrak p}$ over $R_{\mathfrak p}$ we see that (2) $\Rightarrow $ (3).

Finally, assume (3). This means there exists an element $m/f \in M_{\mathfrak p}$ whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$. Then the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$ in $M$ is such that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$. Clearly this means that $\mathfrak p$ contains $I$ and is minimal among the primes containing $I$, i.e., (1) holds. $\square$


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