Minimal support primes of a finite module are associated

The stated AC discharges [L1]'s premise. Choose a prime filtration as in [L1]. Repeatedly applying [L2] to the short exact sequences $0⟶M_{i-1}⟶M_i⟶M_i\mathrm{/}{M}_{i-1}⟶0$ and then using [L3] for $M_i\mathrm{/}{M}_{i-1}\cong R\mathrm{/}{\mathfrak{p}}_i$ gives ${\mathrm{Supp}}_R(M)=V({\mathfrak{p}}_1)\cup \cdots \cup V({\mathfrak{p}}_n)\mathrm{.}$

By step 1.1, the prime $\mathfrak{p}$ contains some ${\mathfrak{p}}_i$. Since ${\mathfrak{p}}_i\in V({\mathfrak{p}}_i)\subseteq {\mathrm{Supp}}_R(M)$, the minimality of $\mathfrak{p}$ in the support forces ${\mathfrak{p}}_i=\mathfrak{p}$. Choose the smallest such index $i$, and choose $m\in M_i$ whose image in $M_i\mathrm{/}{M}_{i-1}\cong R\mathrm{/}\mathfrak{p}$ is nonzero. Then ${\mathrm{Ann}}_R(m)\subseteq \mathfrak{p}$, because any scalar killing $m$ kills its nonzero class in $R\mathrm{/}\mathfrak{p}$. Also ${\mathfrak{p}}_j{M}_j\subseteq M_{j-1}$ for every $j$, so $({\mathfrak{p}}_1\cdots {\mathfrak{p}}_i)m=0$.

For each $j<i$, the minimal choice of $i$ gives ${\mathfrak{p}}_j\mathrm{\ne }\mathfrak{p}$ and hence ${\mathfrak{p}}_j⊈\mathfrak{p}$. Choose $a_j\in {\mathfrak{p}}_j\setminus \mathfrak{p}$, and put $f=a_1\cdots a_{i-1}$. Because $\mathfrak{p}$ is prime and no factor $a_j$ lies in $\mathfrak{p}$, one has $f\mathrm{\notin }\mathfrak{p}$. Since ${\mathrm{Ann}}_R(m)\subseteq \mathfrak{p}$, this implies $fm\mathrm{\ne }0$. Moreover $\mathfrak{p}fm=0$ because $\mathfrak{p}={\mathfrak{p}}_i$ and $({\mathfrak{p}}_1\cdots {\mathfrak{p}}_i)m=0$. If $bfm=0$, then $bf\in {\mathrm{Ann}}_R(m)\subseteq \mathfrak{p}$; as $f\mathrm{\notin }\mathfrak{p}$ and $\mathfrak{p}$ is prime, this forces $b\in \mathfrak{p}$. Therefore ${\mathrm{Ann}}_R(fm)=\mathfrak{p}$, so $\mathfrak{p}\in {\mathrm{Ass}}_R(M)$.

Thus every support-prime minimal by inclusion is associated.