Let $(R,\mathfrak m)$ be a Noetherian local ring. Suppose that all $x$ belonging to $\mathfrak m-\mathfrak m^2$ are zero divisors. Show that all elements of $\mathfrak m$ are zero divisors.
My progress:
My idea is to see that the set of zero divisors is equal to the union of the associated primes of the ideal zero, obtained from the minimal primary decomposition of the ideal zero.
Best Answer
I think it's useful to promote my comments on Pete Clark's answer to a new answer.
We know that the set of zero divisors of $R$ is equal to the union of the associated primes of $R$. Moreover, the set $\operatorname{Ass}(R)$ of associated primes is finite. We have $$\mathfrak m=\mathfrak m^2\cup(\mathfrak m-\mathfrak m^2)\subseteq\mathfrak m^2\cup\bigcup_{\mathfrak p\in\operatorname{Ass}(R)}\mathfrak p.$$ From prime avoidance lemma we learn that $\mathfrak m$ is contained in one of the ideals of the union, so $\mathfrak m\subseteq\mathfrak m^2$ or $\mathfrak m\subseteq\mathfrak p$ for some $\mathfrak p\in\operatorname{Ass}(R)$.
In the first case $\mathfrak m=0$ by Nakayama, and this is a contradiction with the hypothesis which tacitly assumes $\mathfrak m\ne\mathfrak m^2$.
In the second case we get $\mathfrak m\in\operatorname{Ass}(R)$, and therefore there is a nonzero element $x$ of $R$ with $x\mathfrak m=0$, so $\mathfrak m$ consists of zerodivisors, and we are done.