# An equivalent condition for a ring to be Cohen-Macaulay

I want to prove that a local ring $A$ is Cohen-Macaulay if and only if for every $A$-module $M$ we have $\mathrm{grade}\;M+\mathrm{dim}\;M=\mathrm{dim}\;A$.

If that equation holds we just take $M=k$ (the residue field) and we obtain $A$ is Cohen-Macaulay.

Now suppose that $A$ is CM. In general we have

$\mathrm{grade}\;M=\mathrm{inf}_{p\in\mathrm{Supp}\;M}\;\mathrm{ht}\;p$

$\mathrm{dim}\;M=\mathrm{sup}_{p\in\mathrm{Supp}\;M}\;\mathrm{dim}\;A/p$

Now the notes where I am studying conclude saying that the wanted equality follows from the fact that the spectrum of a Cohen-Macaulay ring is bi-dimensional (that I don't know what it means).

It seems to me that I have to use the property that in a CM ring $\mathrm{ht}\;I+\mathrm{dim}\;A/I=\mathrm{dim}\;A$, and apply the two formulas above, but the fact that is bothering me is that in one of them there is a sup and in the other an inf, so I really don't know how to conclude. Any help?

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Let $M$ be a Cohen-Macaulay $A$-module. Bruns and Herzog, Cohen-Macaulay Rings, Theorem 2.1.2(b) shows that $$\mathrm{grade}(I,M)=\dim M-\dim M/IM.$$ When $M=A$ we get $$\mathrm{grade}\ I=\dim A-\dim A/I.$$

Now take $I=\mathrm{Ann}(M)$ and obtain that $$\mathrm{grade}\;M=\dim A-\dim M.$$

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yeah I know this, the fact that I don't understand is why we maximum and the minumum of the formulas that I wrote in the question are reached at the same prime, could you explain me that? – Chris Oct 29 '12 at 13:44
I'm sorry, I totally misunderstood your answer, now I got it – Chris Oct 29 '12 at 22:36

Probably your notes say "equi-dimensional" rather than "bi-dimensional", meaning that every irred. component of Spec $A$ is of the same dimension when $A$ is CM.

Because of this, we find that the the sum of the inf and sup in your formulas is equal to $\dim A$. In fact, this seems to be the point you are confused about (given your last paragraph), so here are some hints:

1. The inf and sup can be replaced by a min and max, since the quantities appearing in both are integers bounded above by $\dim A$ and below by $0$.

2. For any prime ideal $p$ of a catenary Noetherian ring $A$, the sum of ht $p$ and $\dim A/p$ is equal to the maximum of the dimensions of the irred. components containing $p$. Since in the CM case these are all of dimension equal to $\dim A$, you get that for any $p$ there is an equality ht $p + \dim A/p = \dim A$.

3. From 2. you see that max and min are achieved by the same prime ideal $p$ (namely a minimal prime in Supp $M$), and that the sum of the two equals $\dim A$, as required.

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could you tell me why the max and the min are achieved by the same prime ideal? – Chris Oct 29 '12 at 4:08
@Chris: Dear Chris, The sum of the two quantities is constrained to equal a constant (namely $\dim A$). Thus when one is maximized, the other is minimized. Regards, – Matt E Oct 29 '12 at 14:18