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Let $R$ be a regular local ring, $I$ a prime ideal and $J$ an $I$-primary ideal in $R$. Is it true that if $R/I$ is CM then also $R/J$ is CM? This question is in some way the inverse of this one.

flag	asked Mar 24 2011 at 7:54 Blup 115●1●4

Hailong Dao · Accepted Answer · 2011-03-25 20:14:20Z UTC

A useful way to think about this issue is to consider $J=I^{(n)}$, the $n$-symbolic power of $I$, which by definition is the $I$-primary component of $I^n$.

When $R$ is a polynomial rings over $\mathbb C$, this is the ideal consisting of functions vanishing to order at least $n$ on $X = \text{Spec}(R/I)$.

It is then well-known that the depth of $R/I^{(n)}$ can go down. For example, take $I$ generated by the $2\times2$ minors of the generic $2\times 3$ matrix inside the polynomial rings of the $6$ variables, localized at the maximal ideal of those variables. Then $R/I$ is Cohen-Macaulay of dimension $4$, but $R/I^{(n)}$ would have depth $3$ eventually. For more general statements about ideals of maximal minors, see for example Section 3 of:

Powers of Ideals Generated by Weak d-Sequences, C. Huneke, J, Algebra, 68 (1981), 471-509.

EDIT: the example above looks specific, but such examples should be abound. I expect most Cohen-Macaulay ideals which are not complete intersections to give an example (it is known that $R/I^n$, the ordinary powers, are CM for all $n>0$ iff $I$ is a complete intersection). The $2\times 2$ minors gives is a generic situation of non-complete intersection but CM ideal.

A philosophical comment: it is unlikely that Cohen-Macaulayness will be preserved by basic operations on ideals. So if $R/I, R/J$ are CM, we do not expect $R/\sqrt{I}, R/I^n, R/I^{(n)}, R/P$ ($P$ an associated primes), or $R/(I+J), R/IJ$ etc. to be CM.

The reason is that to preserve depth one needs to control the associated primes, and these operations only allow you to control the support. However, finding an explicit example is usually not so obvious.

Laurent Moret-Bailly · Answer 2 · 2011-03-24 13:13:42Z UTC

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Counterexample: $k$ a field, $R=k[[X,Y]]$, $I=(Y)$, $J=(XY, Y^2)$.

answered Mar 24 2011 at 13:13

Laurent Moret-Bailly
7,276●19●29

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But $J=(xy,y^2)$ is not a primary ideal since $xy \in J$, $y \not \in J$ but no power of $x$ is in $J$. – Blup Mar 24 2011 at 13:59

Sorry, my mistake. That was too simple! – Laurent Moret-Bailly Mar 24 2011 at 16:10

roman · Answer 3 · 2011-03-24 11:03:09Z UTC

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I think it's true. R/I being CM means depth(R/I)=dim(R/I). But as R/I is a factor of R/J by nilradical it follows that dim(R/J)=dim(R/I). As R/I is a factorring of R/J we also have depth(R/I)$\leq$depth(R/J). But as dim(R/I)=depth(R/I)$\leq$depth(R/J)$\leq$dim(R/J)=dim(R/I) we actually have equality dim(R/J)=depth(R/J), an so R/J is CM.

answered Mar 24 2011 at 11:03

roman
190●4

Hi roman, a slight problem here is that the depth of a factor ring might go up (take the union of a plane and a line, the depth then is the depth of the line, which is $1$, but the plane, which is a subvariety, thus factor ring, has depth $2$). – Hailong Dao Mar 24 2011 at 16:07

CM for primary ideal

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