Let $(R,\mathfrak m)$ be a Noetherian local ring of positive depth. Then, is it always true that $\mathfrak m$ can be generated by non-zero-divisors ?
I.e., can we find non-zero-divisors $x_1,\dots,x_n \in \mathfrak m$ such that $\mathfrak m=(x_1,\dots,x_n) ?$
Caution: I do not want $x_1,\dots,x_n \in \mathfrak m$ to be a regular sequence (that would be trivially false)!
Best Answer
Since $R$ has positive depth, $\mathfrak{m}$ contains at least one non-zero-divisor. Let $\mathfrak{n}$ be the ideal generated by all the non-zero-divisors contained in $\mathfrak{m}$. Then, $\mathfrak{m}$ is contained in the union of $\mathfrak{n}$ and the associated primes of $0$ (which are finitely many). Since $\mathfrak m$ contains a non-zero-divisor, it is contained in $\mathfrak n$ by the prime avoidance lemma, i.e., $\mathfrak{m}=\mathfrak{n}$. Finally, you can choose a finite system of generators of $\mathfrak{n}$ since $R$ is Noetherian.
You should note that in this proof it is not relevant that $\mathfrak m$ is a maximal ideal or that $R$ is local. You only need that $R$ is Noetherian and $\mathfrak m$ contains a non-zero-divisor.