For which Noetherian local rings, is every associated prime ideal contained in the square of the maximal ideal?

Question

Let $(R,\mathfrak m, k)$ be a Noetherian local ring of depth $>1$. If $a\in R$ is a zero-divisor, i.e. if $ab=0$ for some $0\ne b \in R$, then must it be true that $a\in \mathfrak m^2$? Since the collection of all zero divisors is the union of all associated primes, so in other words I'm asking: Is it true that every associated prime is inside $\mathfrak m^2$?

Answer 1

Try $R=k[[x,y,z]]/xy$. It has depth two, $xy=0$, but $x$ is not in the square of the maximal ideal.

Answer 2

While the general statement is false, it is true that for regular local rings (i.e. $\dim R = \dim_k \frak{m}/\frak{m}^2$), every zerodivisor (and hence associated prime) is contained in $\frak{m}^2$. (The following argument is inspired by Exercise 12.1.B and 12.2.B of Vakil's THE RISING SEA Foundations of Algebraic Geometry.)

Proof. If $(R,\mathfrak{m},k)$ is regular and $a\in \mathfrak{m}$, then we see that $$ \dim_k \frac{\mathfrak{m}/(a)}{(\mathfrak{m}/(a))^2} = \dim_k \frac{\frak{m}}{(a)+\frak{m}^2} = \begin{cases} \dim_k \mathfrak{m}/\mathfrak{m}^2 &\text{if $a\in \mathfrak{m}^2$}\\ \dim_k \mathfrak{m}/\mathfrak{m}^2 - 1 &\text{if $a\in \frak{m}\setminus\frak{m}^2$} \end{cases} $$ On the other hand, by Corollary 11.18 in Atiyah-Macdonald's Commutative algebra, $$ \dim R/(a) = \begin{cases} \dim R &\text{if $a$ is a zerodivisor}\\ \dim R - 1 &\text{if $a$ is not a zerodivisor} \end{cases} $$ We can now prove the contrapositive: if $a\in \frak{m}\setminus \frak{m}^2$, then by regularity $$ \dim R - 1 \leq \dim R/(a) \leq \dim_k \frac{\mathfrak{m}/(a)}{(\mathfrak{m}/(a))^2} = \dim_k \mathfrak{m}/\mathfrak{m}^2 - 1 = \dim R -1, $$ whence $\dim R/(a) = \dim R - 1$ and $a$ is not a zerodivisor. Q.E.D.