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.


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 1, 2010 at 8:41
  • $\begingroup$ 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. $\endgroup$ Commented Aug 1, 2010 at 17:57

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!

  • $\begingroup$ Hi damiano, I have to make the ideal $I$ codimension 1 because the OP assumes that $I$ does not have codimension $n$! $\endgroup$ Commented Aug 1, 2010 at 7:55
  • $\begingroup$ Otherwise, we can just take $I=(a,b)$, and it will be the same as your example. $\endgroup$ Commented Aug 1, 2010 at 7:58

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