# If ideal can be generated by zero divisors, then is the depth of the ideal 0?

Let $R$ be a Noetherian ring and $I$ an $R$-ideal. The number $\operatorname{depth}_I R$ is the length of maximal $R$-regular sequence in $I$. It is well-known that If $\operatorname{depth}_I R = 0$, then any $x \in I$ is a zero divisor. Is the converse true?

If $I$ can be generated by zero divisors, then is $\operatorname{depth}_I R = 0$?

If $I$ is principal then, it is obvious. I believe the answer is yes in general, but I do not know how to prove or disprove this. A simpler question is that

If $x$ and $y$ are zero divisors, then is $x + y$ a zero divisor?

• Consider the direct product of two fields. The two nob-trivial idempotent elements are zero divisors, yet their sum is not. – Mariano Suárez-Álvarez Feb 6 '16 at 5:29
• In that example, the ideal I=R can be generated by zero-divisors. What is its depth? – Mariano Suárez-Álvarez Feb 6 '16 at 5:31
• @MarianoSuárez-Alvarez Thank you very much. That totally makes sense. Do you know an example in the connected case? – Youngsu Feb 6 '16 at 5:44

Let $R=k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$, and $I=(x_1,\dots,x_4)$. Notice that $x_2-x_1\in I$ is a non-zerodivisor. (We have $\dim R=2$, and $\operatorname{depth}R=1$.)
• Thank you very much. Now, I understand it much clearer. I guess the "minimal" example can be "$k[x,y]/(xy)$" with $x+y$. – Youngsu Feb 7 '16 at 0:47