5
$\begingroup$

Let $A=\bigoplus_{\ell\geq 0}A_\ell$ be a finitely generated graded $\mathbb{Z}$-algebra, with $A_0=\mathbb{Z}$, that is free as a $\mathbb{Z}$-module.

If $k$ is a field, then $A\otimes k$ is a finitely generated graded $k$-algebra, and it is a standard fact (the "Hironaka criterion") that $A\otimes k$ is a Cohen-Macaulay ring if and only if it is free as a module over the $k$-subalgebra generated by a homogeneous system of parameters. (A basis for $A\otimes k$ as an algebra over this polynomial subring is a "Hironaka decomposition".)

I have more or less convinced myself (argument below) that there is a similar statement for $A$ itself, to be made in a precise way momentarily. However, my argument seems unnecessarily laborious and indirect, and uses a few ideas I don't feel in complete command of. Meanwhile, it seems to me if the statement is true that it should be standard, though I haven't been able to find it. So,

Is the following statement a standard fact? Can you give a print reference? (Or is it actually false, in which case where in the argument below have I deluded myself?)

Claim (Hironaka criterion over $\mathbb{Z}$): Let $n = \dim A - 1$. Suppose $\theta_1,\dots,\theta_n$ are homogeneous, positive-degree elements of $A$ such that $A$ is finite as a module over $\mathbb{Z}[\theta_1,\dots,\theta_n]$. Then $A$ is Cohen-Macaulay if and only if it is free as a $\mathbb{Z}[\theta_1,\dots,\theta_n]$-module.

Here is why I think it. I have added asterisks ( * ) at places where I am unsure of my logic. Comments are welcome.

First note that since $A$ is module-finite over $\mathbb{Z}[\theta_1,\dots,\theta_n]$, $A/(\theta_1,\dots,\theta_n)$ is module-finite over $\mathbb{Z}$, by right-exactness of the tensor product. Thus $A/(\theta_1,\dots,\theta_n)$ is integral over $\mathbb{Z}$, so it is one-dimensional.

Also note that the finiteness of $A/(\theta_1,\dots,\theta_n)$ implies that the positively graded ideal $A_+$ is nilpotent mod $(\theta_1,\dots,\theta_n)$. (Indeed $A_+^M =0$ for $M$ bigger than the biggest degree occurring in $A/(\theta_1,\dots,\theta_n)$. $A_+$ is prime (as $A/A_+ = \mathbb{Z}$), so it must be the radical of $(\theta_1,\dots,\theta_n)$.

$\Rightarrow$ Suppose $A$ is Cohen-Macaulay. Under this supposition, since $\dim A/(\theta_1,\dots,\theta_n) = \dim A - n$, $\theta_1,\dots,\theta_n$ must be a regular sequence ( * ), and it is contained in the maximal ideals $A_+ +(p)$ for each prime $p\in\mathbb{Z}$, where $A_+$ is the positively graded ideal of $A$. Since these maximals are height $\dim A = n+1$, they accommodate regular sequences of length $n+1$, and since each is generated by its respective $p$ over $A_+$ which is the radical of $(\theta_1,\dots,\theta_n)$, it must be that $p$ is regular in $A/(\theta_1,\dots,\theta_n)$ for all $p$ ( * ). In particular, $A/(\theta_1,\dots,\theta_n)$ is torsion-free as a $\mathbb{Z}$-module. Since it is also finite, this means it is free. Now we can apply a general result (which I separately convinced myself of) that if $\theta_1,\dots,\theta_n$ is a regular sequence consisting of homogenous positive-degree elements in a graded ring $A$ and $B$ is a graded subring satisfying $B\cap (\theta_1,\dots,\theta_n)A = (0)$ (here we are taking $B=\mathbb{Z}$), then a basis of $A/(\theta_1,\dots,\theta_n)$ as $B$-module lifts to a basis of $A$ as $B[\theta_1,\dots,\theta_n]$-module. We conclude $A$ is free as a $\mathbb{Z}[\theta_1,\dots,\theta_n]$-module in our case.

$\Leftarrow$ Suppose $A$ is free as a $\mathbb{Z}[\theta_1,\dots,\theta_n]$-module. Then for $k=\mathbb{F}_p$ or $k=\mathbb{Q}$, $A\otimes_\mathbb{Z} k$ is free as a $k[\theta_1,\dots,\theta_n]$-module. It is of course also finite as $k[\theta_1,\dots,\theta_n]$-module. Also $\dim A\otimes k = n$, because $A$ is assumed free (thus torsion-free) as a $\mathbb{Z}$ module and therefore $p$ is a regular element for each $p$ (thus $A\otimes \mathbb{F}_p = A/pA$ has dimension $\dim A - 1$, and $A\otimes\mathbb{Q} = A_{\mathbb{Z}\setminus\{0\}}$ has dimension the height of $A_+$ which is $n$). Thus $\theta_1,\dots,\theta_n$ is a homogeneous system of parameters for $A\otimes k$ for each of these $k$'s, so since $A\otimes k$ is free as $k[\theta_1,\dots,\theta_n]$-module, we can invoke the Hironaka criterion to conclude $A\otimes k$ is Cohen-Macaulay. Now any field $K$ has a prime subfield some $k$ we've already considered, and since $A\otimes k$ is a finitely generated algebra over $k$, it is Cohen-Macaulay simultaneously with $A\otimes k \otimes K = A\otimes K$. Thus $A\otimes K$ is Cohen-Macaulay for every field $K$. In particular, it is Cohen-Macaulay for $K$ any residue field of $A$. Since $A$ is free and therefore faithfully flat as a $\mathbb{Z}$-module, its Cohen-Macaulayness follows from that of $\mathbb{Z}$ plus that of each fiber $A_\mathfrak{p}/ pA_\mathfrak{p} = A\otimes_\mathbb{Z}\kappa(\mathfrak{p})$ ( * ), where $\mathfrak{p}\in\operatorname{Spec}A$ and $p$ is defined by $(p) = \mathfrak{p}\cap \mathbb{Z}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.