If $(x,y)$ is a regular sequence in a commutative Noetherian ring , then is it true that $(y,x)$ is also a regular sequence ?
Let $R$ be a commutative ring. If $(x,y)$ is a regular sequence in $R$, then I can show that the image of $x$ in $R/yR$ is not a zero divisor and $x S=S$ where $S=ann_R(y)$. Hence if $R$ is Noetherian, then under the further assumption that $x \in J(R)$ [the Jacobson radical of $R$] , it follows by Nakayama lemma that $S=0$ , hence $y$ is not a zero-divisor, hence it follows that $(y,x)$ is a regular sequence. I have no idea what happens if $x \notin J(R)$.
Please help . Thanks in advance
In general, just because $x,y$ is a regular sequence, this does not imply that $y,x$ is also regular.
Consider the example $R = k[x,y,z]/(x-1)z$ and the sequence of elements $x,(x-1)y$. The ideal they generate is $(x,(x-1)y) = (x,y) \not= R$. It's also not hard to see that $x$ is a nonzerodivisor in $R$ and $R/(x) = k[x,y]/(z)$. Here the image of $(x-1)y$ is not a zerodivisor, because $y$ isn't. So $x,(x-1)y$ is a regular sequence.
But $(x-1)y$ is a zerodivisor in $R$, it's killed by $z$, so the sequence in the reverse order is not regular.
It is the case that every ideal generated by a regular sequence is generated by a set of elements that is a regular sequence in any order!
I pulled this example from Chapter 17 of Eisenbud's Commutative Algebra. Check if out if you want to learn more about regular sequences/Koszul complexes, it's a great resource.