Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Doesn't this give a counterexample?… – Matt Feb 11 '11 at 17:45
@Matt: I don't think so. $x,z(1-x)$ and $x,y(1-x),z(1-x)$ are both regular there. – Dev Bappa Feb 11 '11 at 20:47
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thanks. I only looked at the first response to that question. – Dev Bappa Feb 11 '11 at 23:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.