**Proof.**
Choose a filtration as in (3). In Lemma 10.62.5 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 10.63.2 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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