Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $R$ is a Cohen-Macaulay ring. It is well known that if $I$ is an ideal of $R$ generated by $n$ elements, and $I$ has codimension $n$, then $R/I$ is also Cohen-Macaulay.

Now suppose that $I$ does not have codimension $n$, but (the scheme defined by) $R/I$ has several irreducible components, one of which has codimension $n$. Is (the coordinate ring of) that component necessarily Cohen-Macaulay?

Because being Cohen-Macaulay is a local condition, it is clear that the component is generically (in fact everywhere it does not intersect the other components) Cohen-Macaulay, but there is no reason obvious to me why this would extend to the whole component.

share|improve this question
add comment

2 Answers 2

up vote 5 down vote accepted

Hi Alex, I think this fails for $n=2$. Start with a polynomial ring $S$ and height $2$ prime $P$ such that $S/P$ is not Cohen-Macaulay (for example let $P$ be the kernel of the map $S=k[a,b,c,d] \to k[x^4,x^3y,xy^3,y^4]$). Let $a,b$ be a regular sequence in $P$. Let $R=S[t]$ and $I=(ta,tb)$. Then $I$ has height $1$ and is $2$-generated but the components of $I$ is $(t)$ and the components of $(a,b)$, which include $P$. But $R/P$ is not Cohen-Macaulay because $S/P$ isn't.

share|improve this answer
As I commented on damiano's answer, the reason one can not stop at $I=(a,b)$ is because the OP says "$I$ does not have codimension $n$". If we do not care about that, then there is no need for the silly-looking last 3 sentences. –  Hailong Dao Aug 1 '10 at 8:41
The $a,b,c,d$ in the second sentence should perhaps be $A,B,C,D$ to avoid confusions with the third sentence. I don't want to bump the question, so this comment will suffice. –  Hailong Dao Aug 1 '10 at 17:57
add comment

To complement Hailong's answer, here is another way of seeing that an irreducible component of a Cohen-Macaulay scheme need not have special properties. In the example below, the whole scheme is Cohen-Macaulay, so all of its components have the same dimension, unlike in Hailong's example.

Let $X$ be a reduced and irreducible subscheme of projective space $\mathbb{P}^n$ of codimension $c$. Choose $c$ elements $f_1,\ldots,f_c$ in the ideal of $X$ in $\mathbb{P}^n$ with the property that the vanishing set $X'$ of $f_1,\ldots,f_c$ is reduced and has codimension $c$. Thus the scheme $X'$ contains $X$ as a component, it is a complete intersection, and hence it is Cohen-Macaulay. On the other hand, $X$ was essentially arbitrary, so you could have chosen it to be not Cohen-Macaulay!

For a more explicit example, let $X$ be a surface in $\mathbb{P}^4$ with a point that analytically locally looks like the union of two planes at a single point. (This is one of the standard example of a scheme that is not Cohen-Macaulay.) Choose two "general" elements of the ideal of $X$, and let $X'$ be the scheme defined by those two elements. Thus, $X'$ is Cohen-Macaulay and $X$ is a component of $X'$, but the surface $X$ it is not Cohen-Macaulay.

EDIT As Hailong pointed out, I answered a question that is different than what was asked. To answer the initial question, it suffices to argue as above, choosing also a hypersurface $H$ in $\mathbb{P}^n$ not containing $X$. Denote by $h$ an equation for $H$ and replace $f_1 , \ldots , f_c$ by $h f_1 , \ldots , h f_c$. The vanishing set $\overline{X}$ of these equations consists of the union of $H$ and the scheme $X'$ we had before. Thus $X$ is still a component of $\overline{X}$, of codimension $c$ in $\mathbb{P}^n$, the ideal of $\overline{X}$ is generated by $c$ equations, but the component $X$ is (essentially) arbitrary, in particular it need not be Cohen-Macaulay. This should now answer the question that was asked!

share|improve this answer
Hi damiano, I have to make the ideal $I$ codimension 1 because the OP assumes that $I$ does not have codimension $n$! –  Hailong Dao Aug 1 '10 at 7:55
Otherwise, we can just take $I=(a,b)$, and it will be the same as your example. –  Hailong Dao Aug 1 '10 at 7:58
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.