Let $I$ be any ideal in $R=\mathbb C[x,y]$ of height 2. If we know the localization of $I$ at any maximal ideal $m\supseteq I$ is generated by a regular sequence of length two in $R_m$. Is this true that $I$ itself is generated by a regular sequence of length two in $R$?
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$\begingroup$ It's pretty late here, so I hope this isn't completely nonsensical. A Gorenstein ring is Cohen-Macaulay and both $R_m$ and $R$ are local Gorenstein and so local Cohen-Macaulay. For such rings any ideal $I$ has height equal to the depth of $I$ with respect to $I$. See e.g. en.wikipedia.org/wiki/Height_(ring_theory). Also, all regular sequences in $I$ have length equal to the depth of $I$, see en.wikipedia.org/wiki/Depth_(algebra). Now, $R$ is already local, so it seems height = depth = length of longest regular sequence even without this bit about $R_m$. How does that sound? $\endgroup$– David WhiteCommented Aug 10, 2012 at 3:52
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$\begingroup$ What do you mean by '$R$ is already local'? $\endgroup$– J.C. OttemCommented Aug 10, 2012 at 7:49
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$\begingroup$ Could you give the reference for the claim: If $R$ is local then $I$ generated by a regular sequence? $\endgroup$– Pham Hung QuyCommented Aug 10, 2012 at 12:45
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2 Answers
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The quickest way to see this is the following. If $I_{\mathfrak{m}}$ is generated by a regular sequence, using Koszul complex, one gets that $Ext^1(I,R)$ is locally isomorphic to $R/I$ and thus by Chinese remainder theorem, globally isomorphic to $R/I$. Then one checks that the extension corresponding to $1\in Ext^1(I,R)$, say $0\to R\to P\to I\to 0$ has $P$ a projective $R$-module of rank 2 and by Seshadri's theorem $P$ is free. Thus $I$ is 2-generated. Easy to arrange the generators to be a regular sequnce.
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$\begingroup$ Thanks for your reply. Could you explain in details why the identity $1\in Ext^1(I,R)$ corresponds to a ses $0\rightarrow R\rightarrow P\rightarrow I\to 0$ where $P$ is projective? $\endgroup$– XingtingCommented Aug 14, 2012 at 0:49
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$\begingroup$ Let me describe one of may ways of seeing this. By the very nature of it, $P$ is torsion free and hence has depth at least one. That it comes from a generator of $Ext^1(I,R)$ also implies $Ext^1(P,R)=0$. If $P$ is not projective, by Auslander-Buchsbaum formula, $pd P$ must be one. Writing a resolution $0\to A\to B\to P\to 0$ with $A,B$ free, using the fact that $Ext^1(P,R)=0$, one easily checks that $P$ must be projective. $\endgroup$– MohanCommented Aug 14, 2012 at 13:31
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Hi. I'm a little bit confused by the above example. The way I looked at this is through Forster-Swan number.
[Forster] Let $R$ be a Noetherian ring and $M$ a finite $R$-module. Then $M$ can be generated by $b(M)$ elements.
Here $b(M) = 0$ if $M = 0$ and $b(M) =$ sup { $\mu(M_p) + $ dim $R/p$ } where $p \in $ supp$(M)$ if $M \neq 0$.
If we take $R = \mathbb{C}[x,y]$ and $M = I$ as in the question. Then $b(M) = 2$. So $I$ can be generated by two elements.
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$\begingroup$ In this case $b(M)=3$, since $0\in Supp(M)$ where $M=I$. In fact, Forster-Swan does not use the fact you are working over a polynomial ring and in general, you will not be able to do better than 3 for arbitrary rings of dimension 2. $\endgroup$– MohanCommented Aug 11, 2012 at 12:05
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$\begingroup$ Yes. You are right. It is 3 not 2. Thank you for the clarification. $\endgroup$– YoungsuCommented Aug 13, 2012 at 2:33