# Cohen-Macaulay and connected implies equidimensional?

I'm asking for a reality check.

It seems to me that since Cohen-Macaulay rings are locally equidimensional, such a ring is either equidimensional or else disconnected (with different dimensions occurring on different connected components).

But I have not slept enough, so I don't trust my reasoning. Is it sound?

Addendum: Having given this more (and better-slept) thought, it is not at all clear to me. I know that a local Cohen-Macaulay ring is equidimensional, in other words, all minimal primes have the same dimension. (E.g. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry, Corollary 18.11.)

(Edit 2/24/17: The following proof is flawed. See second addendum.)

It follows from this that for an arbitrary (noetherian) Cohen-Macaulay ring $R$, if $\operatorname{Spec} R$ is connected, then all the minimal primes $\mathfrak{p}$ have the same dimension. I.e. $\dim R/\mathfrak{p}$ does not depend on $\mathfrak{p}$. Proof: two minimal primes of differing dimension cannot be contained in the same maximal, by localizing at that maximal and applying the result for local rings. Therefore, sort the minimal primes according to dimension. There are only finitely many minimal primes, so finitely many classes containing finitely many primes each, and one can build a partition of $\operatorname{Spec} R$ into finitely many disjoint closed sets corresponding to the classes in this way. Connectedness implies there is only one class, thus all minimal primes have the same dimension.

Geometrically, this is the statement that the irreducible components of $R$ are all the same dimension.

But I thought equidimensionality also meant that the maximals all have the same height. Here, I don't see the argument. Imagine one maximal that covers all the minimals with height $1$, say, and another maximal that lies above all the minimals with height $2$, with a lot of height 1 primes in between. This yields a connected $\operatorname{Spec}$ since a single closed set containing all the minimal primes will be everything, but if two different closed sets each contain a minimal prime, then they overlap at the maximals.

This scenario is not ruled out by any property I can think of possessed by the poset of primes of a Cohen-Macaulay ring: it is catenary (since it is a ranked poset, ranked by height); the minimals are all dimension 2; etc.

But yet I see numerous references on the internet to the notion that Cohen-Macaulay rings are equidimensional (sometimes without even a connected assumption, which can't be right; e.g. $k\times k[x]$ is Cohen-Macaulay). E.g. here. Perhaps they mean the local case? Or are using a weaker notion of equidimensionality?

Second addendum, 2/24/17: I am no longer convinced by the above argument that the minimal primes in a Cohen-Macaulay ring with a connected spectrum must have the same dimension. It is the case that the minimal primes under a given maximal have the same-length saturated chains to that maximal, but it doesn't follow that they have the same dimension: perhaps one of them is also under another maximal with a greater height, while the other is not.

On the other hand: if the Cohen-Macaulay ring with the connected spectrum happens to be integral over an equidimensional Cohen-Macaulay subring (which is guaranteed, e.g. for finite-type $k$-algebras by Noether normalization, per conv. with MooS below her/his answer), then I have convinced myself it is equidimensional. Here is why I think so:

Let the Cohen-Macaulay ring be $R$ and let the equidimensional Cohen-Macaulay subring be $S$; say its dimension is $d$.

Since $S$ is Cohen-Macaulay, the result quoted above for local rings, together with the fact that Cohen-Macaulay rings are catenary, implies that the lengths of any two saturated chains of primes from a given maximal to two minimals under it are equal. Since $S$ is equidimensional, this common length must be $d$ for every maximal. Thus, the length of any saturated chain between any minimal prime and any maximal is $d$. It follows from a second application of catenaryness that for any prime ideal of $S$, and any saturated chain from a minimal to a maximal containing this prime, the position of this prime in this chain depends only on the prime and not on the chain.

Now consider any saturated chain of primes between a minimal prime $\mathfrak{p}$ and a maximal $\mathfrak{m}$ in $R$. Say its length is $\ell$. Intersecting with $S$, one obtains a chain of primes, and due to integrality, it has the following three properties: (1) it is saturated; (2) its "top end" is a maximal; (3) it is length $\ell$. (1) and (2) are due to going-up and (3) is due to incomparability. (Edit 2/24/17: I'm no loger convinced going-up implies (1). Grr! See third addendum.)

Let the $h(\mathfrak{p})$ be the height of $\mathfrak{p}\cap S$ in $S$. By the above discussion, $h(\mathfrak{p}) = d - \ell$; thus $\ell$ doesn't depend on $\mathfrak{m}$ or the choice of chain, but only on $\mathfrak{p}$.

Connectedness of the spectrum (at least for noetherian rings) is equivalent to connectedness of the bipartite graph with vertex classes the minimal and maximal primes, and edges indicating containment. (Proof: two closed sets $V(\mathfrak{p}), V(\mathfrak{q})$ of the spectrum, where $\mathfrak{p},\mathfrak{q}$ are minimal primes, meet iff there is a maximal containing both $\mathfrak{p},\mathfrak{q}$. So if the graph is disconnected, there is a partition of these finitely many closed sets $V(\text{minimal prime})$, which cover the spectrum, into disjoint classes, and this disconnects the spectrum. Conversely, if the spectrum is disconnected, there are two disjoint nonempty closed sets; they must both meet the minimal primes and contain everything above the ones they meet, and then the way they partition the minimal primes also partitions the graph.) Thus, between any two minimal primes $\mathfrak{p},\mathfrak{q}$, there is a path in the graph

$$\mathfrak{p} = \mathfrak{p}_1 \subset \mathfrak{m}_1\supset \mathfrak{p}_2\subset \mathfrak{m}_2\supset \dots\subset\mathfrak{m}_r\supset \mathfrak{p}_{r+1} = \mathfrak{q}$$

Because $R$ is Cohen-Macaulay, by the same logic as for $S$ above (i.e. by localizing at $\mathfrak{m}_i$) we conclude that the lengths $\ell_i$ and $\ell_i'$ of saturated chains between $\mathfrak{m}_i$, and $\mathfrak{p}_i$ and $\mathfrak{p}_{i+1}$ respectively, are equal, for each $i$. Thus $h(\mathfrak{p}_i)$ and $h(\mathfrak{p}_{i+1})$ are equal. And thus $h(\mathfrak{p})$ and $h(\mathfrak{q})$ are equal. Thus $h$ is a constant function on the minimal primes of $R$.

Now take any minimal prime of $S$. By lying-over, there is some prime of $R$, say $\mathfrak{p}^*$, lying over it, which must be minimal by incomparability, and we have $h(\mathfrak{p}^*)=0$. Therefore, $h$ is identically zero on the minimal primes of $R$. Thus for any saturated chain in $R$ from a minimal $\mathfrak{p}$ to a maximal $\mathfrak{m}$, of length $\ell$, we have $0=h(\mathfrak{p}) = d - \ell$, and we conclude $\ell=d$.

Therefore, all maximals have height $d$ and all minimals have dimension $d$, i.e. $R$ is equidimensional of dimension $d$.

Third addendum, 2/24/17: There is a soft spot in the proof of the second addendum as well. I found another unjustified assumption I made probably due to being overly familiar with the good behavior of the coordinate rings of varieties. Going-up doesn't imply by itself that the intersection of a saturated chain in $R$ with $S$ yields a saturated chain in $S$, although it may well be true in the present context for other reasons. I asked a followup question on this point here.

Fourth addendum, 2/26/17: Given the track record at this point, there's no reason to expect me not to find another mistake; however, I believe I have proven this variant of the statement in the second addendum:

If a Cohen-Macaulay ring has a connected spectrum and is finite (strengthened from integral) over a subring that is an integrally closed, equidimensional, catenary noetherian domain, then it is equidimensional.

This OP is already too long without me adding another detailed proof, but it very loosely follows the pattern of the argument in the 2nd addendum, i.e. argues that all maximal chains in $S$ have length $d$ and then shows $h(\mathfrak{p})=0$ using connectedness. Here is a reasonably complete outline:

I. Show all maximal chains in $S$ have the same length $d$ using equidimensionality + catenariness.

II. Consider the set $W_0$ of primes of $R$ whose intersection with $S$ is zero.

(a) They are all minimal and dimension $d$.

(b) Fixing one of them $\mathfrak{p}$, and a target maximal $\mathfrak{m}$ containing it, construct a chain of length $d$ to $\mathfrak{m}$ by (b1) passing temporarily to $R/\mathfrak{p}$ which is a domain containing $S$; then passing to their fraction fields; then forming normal closure of $\operatorname{Frac}(R/\mathfrak{p}) /\operatorname{Frac}S$ and taking $S$'s integral closure $B$ in it; (b2) invoking going-up to construct a chain of length $d$ in $B$ with top term lying over $\mathfrak{m}\cap S$, and and lying-over to find another prime in $B$ lying over $\mathfrak{m}$; (b3) using a Galois automorphism to move the chain til its top term is the prime we found lying over $\mathfrak{m}$; (b4) intersecting with $R$ yielding the desired chain.

(c) By Cohen-Macaulayness, a maximal $\mathfrak{m}$ of $R$ that contains one of the $\mathfrak{p}$'s in $W_0$ (and is therefore height $d$) can't also contain a minimal prime $\mathfrak{q}$ with $h(\mathfrak{q}) >0$. Thus if there are any such minimal primes, the graph described in the 2nd addendum is disconnected. This contradicts the fact that the spectrum of $R$ is connected, so there are no such primes.

Therefore all minimals are in $W_0$, so are dimension $d$ by (a), and all maximals contain one of these, so are height $d$ by (b). Conclude equidimensionality.

Fifth addendum, 3/13/17: Fourth addendum holds up I believe, but argument in (b) can be simplified by replacing the Galois theory just by passing to $R/\mathfrak{p}$ and then invoking the going-down theorem for $S\subset R/\mathfrak{p}$.

Consider $R=k[x,y]$ and the multiplicative system $S = R \setminus (P \cup Q)$ with $P=(x,y)$ and $Q=(x-1)$. $S^{-1}R$ is not equidimensional since its maximal ideals are $P$ and $Q$ of height $2$ and $1$ respectively.

But $S^{-1}R$ is clearly Cohen-Macaulay, since all its localizations are regular local rings.

When you read at some point that the author concludes equidimensionality from the Cohen-Macaulay property, then he probably either works in the local case or in the case of a finite type $k$-algebra. Or he simply made a mistake.

• Is the argument for finite type $k$-algebras sufficiently quick to give it here? (My situation is a finite type $\mathbb{Z}$-algebra and I'd like to see if it works in that situation.) Feb 22, 2017 at 19:27
• In a finite type $k$-algebra, all maximal ideals have always the same height, no Cohen-Maucaulay property is needed for that. That is basically the content of Noether Normalization.
– MooS
Feb 22, 2017 at 19:28
• I am not sure about the finite type $/ \mathbb Z$ case at the moment.
– MooS
Feb 22, 2017 at 19:46
• Sorry, for some reason I assumed you are talking about domains. For non-domains it should still be true if you have the Cohen-Macaulay property, since you can divide out a minimal prime ideal without changing the dimension and thus reduce to the domain case.
– MooS
Feb 22, 2017 at 20:28
• One should maybe phrase it like this: For a finite type $k$-algebra, the properties 'all minimal primes have the same dimension' and 'all maximal primes have the same codimension/height' are equivalent. Sufficient conditions for the first property are for example being a domain or being Cohen-Mac with connected spectrum. Sufficient for the second property is of course local. So these 3 properties are all sufficient for equidimensional in the finite type $/k$ case. Now onto the $/\mathbb Z$ case :)
– MooS
Feb 22, 2017 at 20:51