# Dimension definition

Consider the definition of the topological dimension of a topological space $$X$$ as the maximum of an increasing chain of closed irreducible subsets of $$X$$.

Is dimension a topological concept?
I mean here, If I define the dimension of a ring to be its Krull dimension, and then use this in schemes, am I still talking about the topological concept?
Now, if I have a manifold, and declare its dimension to be the dimension of the Euclidian spaces it resembles on a small scale, is this equivalent to its dimension as a topological space?
Hopefully, my questions make sense.

## 2 Answers

You're dealing with two notions of dimension which only coincide in a degenerate sort of situation. I would call the first notion the Krull dimension of a topological space (the supremum of lengths of chains of irreducible closed subsets), because, indeed, if $A$ is a ring, then its Krull dimension in the sense of commutative algebra coincides with the Krull dimension of $\mathrm{Spec}(A)$.

Krull dimension is not so interesting for Hausdorff spaces like manifolds. Indeed, in a Hausdorff space, the only irreducible closed subsets are singletons, so all Hausdorff spaces have Krull dimension zero. Thus all manifolds have Krull dimension zero. In particular $\mathbf{R}^n$ has Krull dimension zero for all $n\geq 1$. But as a manifold its dimension is $n$. So the only time Krull dimension coincides with the usual notion of dimension for a manifold is if the manifold is zero-dimensional.

For $$X$$ a topological space, let's define $$\dim_{\text{top}} X$$ to be the maximal $$n$$ so that there exists a chain of subsets

$$X_0 \subsetneq X_1 \subsetneq \dots \subsetneq X_n$$

where each $$X_i$$ are irreducible and closed. One then proves that

$$\dim R = \dim_{\text{top}}X$$ when for $$X$$ an algebraic variety over a field $$K$$ with coordinate ring $$R$$ (more generally, if $$X = \text{Spec} (R)$$ is an affine scheme).

The fact that the Krull dimension and topological dimension coincide boils down to the observation that chains of prime ideals in $$R$$ correspond to chains of irreducible closed subsets in $$X$$. Said otherwise, the structure of ideals in the coordinate ring gives topological information about the associated variety -- perhaps not terribly surprising given that they are used to define the topology on the space :)

In the case of a manifold, the topological dimension is badly behaved and does not agree with the Euclidean dimension. Any closed set that is not a point is reducible, so the Krull dimension is always zero!

• Thank you so much, very clear. I accepted you answer, but can I accept more than one answer? . I am not so knowledgeable about this.
– user65304
Commented Aug 16, 2015 at 20:21
• @abu3ttallah you can only accept one answer, but you can upvote any and all answers that you find helpful :) Commented Aug 16, 2015 at 20:22