# What is the "dimension" of a locally ringed space?

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?

• Why does this have so many upvotes? (Ծ‸ Ծ) Oct 29 '20 at 14:17
• @AlexYoucis I am just as surprised as you are Oct 29 '20 at 15:36
• One potential reason: it became a hot network question. You can see this on the post timeline. Oct 29 '20 at 18:06

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.

• 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.) 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? 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. Oct 29 '20 at 18:50
• In fact, the construction in this post 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