Recall that in a commutative ring $A$ an ordered pair of elements (a,b) is said to form a regular sequence if the ideal $\langle a,b\rangle $ is strictly included in $A$ ,if $a$ is not a zero-divisor in $A$ and if the class of $b$ is not a zero-divisor of $A/\langle a\rangle$.  

A friend of mine has asked me if in that case we can conclude that $\langle b,a\rangle $ is also a regular sequence *under the assumption that $A$ is a noetherian domain*.  

The answer is known to be yes for a *local* noetherian ring $A$, even it is not a domain 


[Since I couldn't answer his question, I suggested to my friend that he ask here but he prefers that I do that]



Does a regular pair of elements in a noetherian domain remain regular if their order is switched?
The only part to be shown is that $a$ is not a zero-divisor on $A/(b)$. Consider some $s\in A$ such that $as\in(b)$, say $as=bt$. Since $b$ is not a zero-divisor on $A/(a)$, we conclude that $t$ maps to zero in $A/(a)$, i.e., $t=au$. Since $A$ is a domain, it follows that $s=bu$, so $s$ maps to zero in $A/(b)$.


