I was trying to answer this question - whether a subsequence of a regular sequence is regular in a Noetherian ring which is not local. In the local case, regular sequences can be permuted and so a subsequence can be considered to be the initial subsequence of a regular sequence and hence regular.

Let's consider the following case. Let $R$ be a Noetherian ring which is not local. Let $x_1,x_2,x_3$ be a regular sequence. Is $x_1,x_3$ regular?

Now, $x_1$ is a nonzerodivisor on $R$. So it's really a question of whether $x_3$ is a nonzerodivisor on $R/(x_1)$. Suppose not, then there is an element $y\in R\setminus (x_1)$ s.t. $x_3 y\in (x_1)R$. But since, $x_3$ is a nonzerodivisor on $R/(x_1,x_2)$, we must have $y\in (x_1,x_2)$. So we may assume, $y\in (x_2)$. Write, $y=rx_2$. Then, $x_3 rx_2 \in (x_1)$. But $x_2$ is a nonzerodivisor on $R/(x_1)$. So, $rx_3\in (x_1)$. I am stuck here. Can someone help with the proof or know a counterexample? Thanks.

  • $\begingroup$ Doesn't this give a counterexample? math.stackexchange.com/questions/12408/… $\endgroup$ – Matt Feb 11 '11 at 17:45
  • $\begingroup$ @Matt: I don't think so. $x,z(1-x)$ and $x,y(1-x),z(1-x)$ are both regular there. $\endgroup$ – Dev Bappa Feb 11 '11 at 20:47

I cannot answer in a comment (not enough rep), but what is wrong with Akhil's reply? $(x-1)y$ is clearly a subsequence of $x,(x-1)y$. You could just as well take $R=k[x,y,z,w]/(x-1)z$, $x_1=w$, $x_2=x$, $x_3=(x-1)y$. Then, $x_1,x_2,x_3$ is regular, while $x_1,x_3$ isn't - which is just what you were trying to prove.

  • $\begingroup$ Thanks. I only looked at the first response to that question. $\endgroup$ – Dev Bappa Feb 11 '11 at 23:01

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