# Finding an ideal such that quotient is Cohen-Macaulay

Let $R$ be a commutative local Noetherian ring which is not a domain and not Cohen-Macaulay. Can we find an ideal $I$ in $R$ such that $R/I$ is Cohen-Macaulay, and $\dim R/I=\dim R$?

I don't think the answer is true in general. Let $(A,m)$ be a Noetherian local domain which is not Cohen-Macaulay. Let $R$ be the localization of $A[x]/(x^2)$ at $(m, x)$. Then $(x)$ is the unique minimal prime with $\dim R = \dim R/(x)$, but $R/(x) \cong A$ is not Cohen-Macaulay.

Let $I$ be an $R$-ideal such that $\dim R/I = \dim R$. Write $p = (x)$. Then Min$(R/I) = \{ p \}$ since $\sqrt{0} = p$. Suppose that $R/I$ is Cohen-Macaulay. Then Min$(R/I) =$ Ass$(R/I)$. Hence $I$ is an $p$-primary ideal. Observe that $R_p$ is a localization of $K[x]/(x^2)$ where $K$ is the field of fraction of $A$. Thus $R_p$ is a principal ideal ring. In particular, the only non zero proper ideal is $p_p$. This shows that $I_p = p_p$. Therefore, $I = p$ since $I$ is $p$-primary. This is a contradiction.

Let $R = \{f(1,1) = f(-1,-1)\} \subset k[x,y]$. So $\mathrm{Spec} R$ looks like a surface passing back through itself at an isolated point. The singular point is not Cohen-Macaulay since it looks like two planes meeting at a point (analytically).

Now let $S = R \oplus R$, so $\mathrm{Spec} S$ is the disjoint union of two copies of of this surface. There are no Cohen-Macaulay, dimension 2 closed subsets of $\mathrm{Spec} S$!

Edit: you asked for a local ring, so instead take the union of two copies of the surface meeting at the singular point, and take the local ring of the point.

• by the way, is $R$ a finitely generated $k$-algebra?
– Hans
Jan 24 '15 at 8:54
• Yes. It should be possible to obtain it explicitly (in fact, a compactification of it) by taking the degree-3 embedding $\phi: \mathbb{P}^2 \to \mathbb{P}^{9}$, choosing a point $p$ on the line connecting $\phi(1,1)$ to $\phi(-1,-1)$, and projecting from $p$. Jan 24 '15 at 10:01
• @Hans: My last comment isn't exactly correct -- that construction creates two singularities (which is still fine to answer the original question posed here). But in any case, a generating set for $R' = \{f(0,0) = f(1,1)\}$ (which is easier to calculate) is $\{x-y,y^2-y,y^3-y,x^2-x\}$ (and if you want, $x^3-x$ to make it easier to prove that these generate $R'$.) Jan 25 '15 at 19:07
• I see, thanks..
– Hans
Jan 27 '15 at 6:47