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Consider the following properties of scheme $X$:

A: $X$ is of finite type over $\mathbb{Z}$

B: $X$ is Noetherian

C: $X$ is of finite Krull dimension

What implications are there between these three? I believe that B and C are independent of each other (although I can't find a reference right now), and it follows from EGA I, 6.3.7 that A implies B. But does A imply C?

(Apologies if this question is "trivial", but I'm not an expert in algebraic geometry.)

As an aside, I would also be interested if any of these properties can be related to some notion of cohomological dimension (not sure what kind of topologies would be relevant for this).

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(By the way, your question -- which is more of a commutative algebra question than an algebraic geometry question -- is a standard one. It is certainly not trivial: indeed, the first example of a Noetherian ring of infinite Krull dimension is due to Nagata, and is one of the most famous examples in the entire subject.) –  Pete L. Clark Nov 7 '12 at 9:23
Dear Anon, Grothendieck's vanishing theorem says that the cohomology of a sheaf on an $n$-dimensional Noetherian space vanishes in dimensions above $n$. I think this gives the relationship to cohomological dimension that you are asking about. Regards, –  Emerton Nov 8 '12 at 4:16

2 Answers 2

An example of a noetherian ring of infinite dimension can be found in Nagata's Local Rings, Appendix A1, Example 1.

Edit: An interesting generalisation of Nagata's construction yielding noetherian rings of infinite dimension whose maximal ideals have prescribed heights was given by Fujita in his article Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181–185.

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(True, of course.) And yet a Noetherian local ring must have finite Krull dimension! –  Pete L. Clark Nov 7 '12 at 9:19

A implies B. True, as you said, because a finitely generated ring is Noetherian, and $X$ is glued from finitely many spectra of such.

A implies C. True (argument as above).

B implies A. False, e.g. $X = \mathrm{Spec }\ \mathbb{Q}$.

B implies C. False (I believe). There are rings $R$ whose spectrum is homeomorphic to the topological space $\{1, 2, \ldots \}$ with open sets $\{n, n+1, \ldots\}$, which is Noetherian but of infinite Krull dimension. I think something like $\mathrm{Spec }\ k[x_1, x_1 x_2, x_1 x_2 x_3, \ldots]$ should work, but I didn't check the details. EDIT. This is nonsense - see the comments below and Fred Rohrer's answer.


C implies B. False, e.g. $X = \mathrm{Spec}\ k[x, x^{1/2}, x^{1/3}, \ldots]$.

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Dear @Piotr, the ring $k[x_1,x_1x_2,x_1x_2x_3,...]$ is indeed of infinite dimension, but not noetherian, since its ideal generated by $x_1$, $x_1,x_2$, $x_1x_2x_3$,... is not of finite type. –  Fred Rohrer Nov 7 '12 at 7:30
Well, $x_1$, $x_1$, $x_2$ in my previous comment should obviously read $x_1$, $x_1x_2$. –  Fred Rohrer Nov 7 '12 at 7:40
Dear @Fred, isn't this ideal just generated by $x_1$, or am I missing something? –  Piotr Achinger Nov 7 '12 at 16:19
Re B implies C : By definition, a noetherian scheme $X$ is covered by finitely many spectra of noetherian rings. If $X$ has infinite Krull dimension, then one of these noetherian rings must have infinite Krull dimension. Thus we need a Nagata-like counterexample. –  François Brunault Nov 7 '12 at 21:54
I think your ring $k[x_1,x_1 x_2,x_1 x_2 x_3,\ldots]$ is abstractly isomorphic to $k[X_1,X_2,X_3,\ldots]$. –  François Brunault Nov 8 '12 at 6:56

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