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.