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.



Irreducible components of quotients of Cohen-Macaulay rings of the "correct" dimension
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. 


