Let $R$ the polynomial ring in $n$ variables with complex coefficients and $I$ an ideal of $R$. Is it true that if $R/I$ is CM also $R/J$ is CM (where $J$ is the radical of $I$)? Is there a relations between a resolution of $R/J$ and one of $R/I$? What if I suppose that $proj.dim(R/I)=2$?
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$\begingroup$ Why projective dimension 2? Hartshorne's example $k[s^4,s^3t,st^3,t^4]$ works in char $p>0$. $\endgroup$– Hailong DaoCommented Nov 27, 2010 at 17:12
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$\begingroup$ because I'm looking at a case in which the projective dimension is 2. Sorry, I'm not confident with Hartshorne's example....but I work in char p=0. $\endgroup$– Michele TorielliCommented Nov 27, 2010 at 17:19
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$\begingroup$ Michele: I doubt it will be true in projective dimension 2. If you have a specific situation, you should post the details separately. This is not easy stuff to answer without knowing full details. $\endgroup$– Hailong DaoCommented Nov 27, 2010 at 18:07
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$\begingroup$ sorry, you are right. In my case I is the jacobian ideal of a polynomial f such that R/I is CM of codim 2 and hence proj.dim 2 $\endgroup$– Michele TorielliCommented Nov 27, 2010 at 18:16
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2 Answers
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It is not true, but the example is not easy to find $I = (x_2^2-x_4x_5,x_1x_3-x_3x_4, x_3x_4-x_1x_5)$!
answered Nov 26, 2010 at 15:07
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$\begingroup$ ok, is it true adding some restriction on I? $\endgroup$ Commented Nov 26, 2010 at 15:20
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$\begingroup$ The only thing that jumps to my mind is hypersurfaces. In general, there's probably not much you can say. For example, many many varieties are locally set theoretic complete intersections (ie, the radicals are cut out by a regular sequence locally) but don't satisfy any nice properties themselves. Maybe Long will have some more suggestions? $\endgroup$ Commented Nov 26, 2010 at 15:41
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$\begingroup$ what if R is local?like power series? $\endgroup$ Commented Nov 26, 2010 at 16:07
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$\begingroup$ Michele: the same example works in the power series rings $k[[x_1,\cdots,x_5]]$. In addition to what Karl says, if $I$ is monomial, then what you say is true. $\endgroup$ Commented Nov 26, 2010 at 16:24
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$\begingroup$ With respect to what I said, it should be "(ie, they are cut out by regular sequences up to radical)." $\endgroup$ Commented Nov 26, 2010 at 18:22
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Yes. From Eisenbud's Commutative Algebra: a ring $S$ is Cohen-Macaulay iff all the maximal ideal $m$ of $S$ satisfies codim($m$) = depth($m$). Now, the maximal ideals of $R/J$ are the same as $R/I$ and their depths and codimensions are the same as well.
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$\begingroup$ so it seems that also if R/J is CM then R/I is CM, but this is false if I'm not wrong. $\endgroup$ Commented Nov 26, 2010 at 14:33
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$\begingroup$ You are right. I guess you also need to observe to things: 1) a maximal $R/I$-sequence is also a maximal $R/J$ sequence, so that depth($R/J$) $\geq$ depth($R/I$), and 2) depth of any maximal ideal is less than or equal to its codimension. $\endgroup$– pinakiCommented Nov 26, 2010 at 14:58
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1$\begingroup$ @auniket: I don't think that a $R/I$-sequence always becomes $R/J$-sequence. $\endgroup$ Commented Nov 26, 2010 at 16:26