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*}

Proposition 10.62.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following sets of primes are the same:

  1. The minimal primes in the support of $M$.

  2. The minimal primes in $\text{Ass}(M)$.

  3. For any filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_{n-1} \subset M_ n = M$ with $M_ i/M_{i-1} \cong R/\mathfrak p_ i$ the minimal primes of the set $\{ \mathfrak p_ i\} $.

Proof. Choose a filtration as in (3). In Lemma 10.61.5 we have seen that the sets in (1) and (3) are equal.

Let $\mathfrak p$ be a minimal element of the set $\{ \mathfrak p_ i\} $. Let $i$ be minimal such that $\mathfrak p = \mathfrak p_ i$. Pick $m \in M_ i$, $m \not\in M_{i-1}$. The annihilator of $m$ is contained in $\mathfrak p_ i = \mathfrak p$ and contains $\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_ i$. By our choice of $i$ and $\mathfrak p$ we have $\mathfrak p_ j \not\subset \mathfrak p$ for $j < i$ and hence we have $\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_{i - 1} \not\subset \mathfrak p_ i$. Pick $f \in \mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_{i - 1}$, $f \not\in \mathfrak p$. Then $fm$ has annihilator $\mathfrak p$. In this way we see that $\mathfrak p$ is an associated prime of $M$. By Lemma 10.62.2 we have $\text{Ass}(M) \subset \text{Supp}(M)$ and hence $\mathfrak p$ is minimal in $\text{Ass}(M)$. Thus the set of primes in (1) is contained in the set of primes of (2).

Let $\mathfrak p$ be a minimal element of $\text{Ass}(M)$. Since $\text{Ass}(M) \subset \text{Supp}(M)$ there is a minimal element $\mathfrak q$ of $\text{Supp}(M)$ with $\mathfrak q \subset \mathfrak p$. We have just shown that $\mathfrak q \in \text{Ass}(M)$. Hence $\mathfrak q = \mathfrak p$ by minimality of $\mathfrak p$. Thus the set of primes in (2) is contained in the set of primes of (1). $\square$


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