Let $(X,\mathscr{O}_X)$ be a locally ringed space. If it is a scheme, the natural notion of dimension is the dimension of the subjacent topological space (the size of the biggest chain of irreducible closed subsets). But if $X$ is a manifold, I think that the natural notion of dimension is perhaps the dimension of the Zariski tangent space.

Is there a "good" notion of dimension in locally ringed spaces? If so, how does this notion relates to the dimension of the subjacent topological space, the dimension of the Zariski tangent space and the Krull dimension of the stalks?

Arctic Char
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1 Answers1


I think the answer is no, because "all locally ringed spaces" is a really broad assortment of objects - it contains too many different types of things for only one concept to really get everything done. This isn't really a proof, though, just an explanation of how most common options do not work. (If you can come up with some axioms you'd want dimension to satisfy, perhaps you can edit them in to your post and we'll see about getting a real proof together.)

Purely algebraic notions defined only in terms of data from the local rings can't be compatible with our topological expectations: given any local ring $R$, the one-point space with $R$ as its sheaf of rings gives a locally ringed space. In any morally upright notion of dimension, this space would have dimension zero: it's a point! But this means that any invariant of local rings you pick to give "dimension" should return zero on all local rings, which isn't great.

Purely topological notions don't work across all of things that can be considered locally ringed spaces either. Krull dimension doesn't work for Hausdorff spaces as every irreducible closed set is a singleton, so we always get dimension 0. All the usual notions of topological dimension (Lebesgue covering, small inductive, large inductive) all fail on schemes with a single closed point because any open cover of such a scheme must contain the whole space. This immediately implies that any such notion of dimension must return zero.

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  • Just as a last bit of hope, do you happen to know what is the Krull dimension of the stalks of a manifold? If it is equal to the dimension of the manifold, then this notion works for schemes and manifolds. (Would it also work for formal schemes and analytic spaces? I know nothing about these spaces.) – Gabriel Oct 29 '20 at 11:19
  • Actually, I would argue that the dimension of a locally ringed point should exactly be the dimension of the local ring it carries. Nonetheless, I tend to agree with the general purport of your answer. – Ben Oct 29 '20 at 12:11
  • @Ben, could you explain it a little more? This surely is the good notion of dimension in the context of schemes. But does it also work for manifolds? And what about formal schemes / analytic spaces and etc? – Gabriel Oct 29 '20 at 12:21
  • @Gabriel actually, taking $\dim R$ to be the dimension of $X=(\{pt\},R)$ will always give 0 if $X$ is a scheme: in a scheme, every point has a neighborhood isomorphic as an LRS to an affine scheme, which says $X$ itself must be an affine scheme and thus $R$ has krull dimension zero. As far as smooth functions, certainly every closed set determines a prime ideal, and for any manifold of dimension at least 2 you can consider something like $I_t = \{ f\in C^\infty(\Bbb R^2) \mid f(r\cos(\theta),r\sin(\theta))=0 \ \ \forall |\theta| \leq t\}$ and then localize at the origin. – KReiser Oct 29 '20 at 18:50
  • Maybe I don't understood exactly what you said but I think this is not quite what I meant. What I am asking is if the Krull dimension of the ring of germs of a smooth manifold coincides with the dimension of the manifold or not. – Gabriel Oct 29 '20 at 20:05
  • The idea was to show that for any $n$, one could pick a sequence $t_1,t_2,\cdots,t_n$ and show that $I_{t_n}\subset I_{t_{n-1}}\subset \cdots \subset I_{t_1}$, which would demonstrate that the krull dimension of a local ring of smooth functions on a manifold was infinite, but I now realize that these ideals are not even prime, oops. In fact, I now kind of suspect that the Krull dimension of the ring of germs of a point on a smooth manifold is dimension zero? (Prime ideals should correspond to locally irreducible sets through the point up to equiv, but such things are just the point itself.) – KReiser Oct 29 '20 at 20:34
  • @KReiser I revised the proof that prime ideals of $A$ correspond to irreducible closed subsets of $\operatorname{Spec} A$ and I really don't see how we could adapt the proof to have something like you said. – Gabriel Oct 29 '20 at 20:41
  • Taking Taylor expansion defines a surjective local ring homomorphism $\mathcal C_0^\infty(\mathbb R^n)\to\mathbb R[[x_1,\dots,x_n]]$, so the dimension is at least $n$. But the kernel in enormous. That itself doesn’t prove anything, but there is a wild zoo of non-analytic functions, so I would be very much surprised if it would be finite-dimensional. – Ben Oct 29 '20 at 21:19
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    In fact, the construction in this post https://math.stackexchange.com/a/2078897 works after minor modifications also in the smooth case, providing an infinite chain of prime ideals. – Ben Oct 30 '20 at 15:57