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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]

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1 Answer 1

up vote 7 down vote accepted

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 $A/(b)$.

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Strange, my friend ( a great algebraist who definitely knows what a quotient ring is!) told me he knew a counter-example due to Dieudonné in the case $A$ was a non-noetherian domain. He must have misremembered. Anyway your answer is obviously correct and optimal: thanks a lot. –  Georges Elencwajg Oct 27 '11 at 16:24
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According to Matsumura, there is an example by Dieudonne' in nagoya math j. 1966, pp. 355-356, where A is non noetherian and local but not a domain. –  roy smith Oct 27 '11 at 16:26
    
Why is this the only part to be shown? We also need to show that $b$ is not a zero-divisor on $A$. That, it seems to me, is the hard part. –  Steven Landsburg Oct 27 '11 at 16:28
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Steven, that bothered me too, but it seems A is a domain by hypothesis, so there is no zero divisor. –  roy smith Oct 27 '11 at 16:30
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Update: I just phoned my friend to tell him about a-fortiori's solution. He is very happy and told me he would stop looking for counterexamples... –  Georges Elencwajg Oct 27 '11 at 16:33

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