Lemma 10.66.3. For a reduced ring the weakly associated primes of the ring are the minimal primes.

Proof. Let $(R, \mathfrak m)$ be a reduced local ring. Suppose $x \in R$ is an element whose annihilator has radical $\mathfrak m$. If $\mathfrak m \not= 0$, then $x$ cannot be a unit, so $x \in \mathfrak m$. Then in particular $x^{1 + n} = 0$ for some $n \geq 0$. Hence $x = 0$. Which contradicts the assumption that the annihilator of $x$ is contained in $\mathfrak m$. Thus we see that $\mathfrak m = 0$, i.e., $R$ is a field. By Lemma 10.66.2 this implies the statement of the lemma. $\square$

Comment #3771 by Laurent Moret-Bailly on

Typo in proof: "the annihilator of $\mathfrak{m}$" should be "the annihilator of $x$".\ref{}

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