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I'm using this definition for the (Krull) dimension of a topological space $X$ and (Krull) dimension at a point $x\in X$. In general, given a topological space $X$, one always has $$ \dim X=\max\{\dim T\mid T\subset X\text{ is an irreducible component}\}. $$ If $X$ is a scheme locally of finite type over a field, one has $$ \dim_x X=\max\{\dim T\mid T\subset X\text{ is an irreducible component passing through }x\}. $$ (See 0A21(5).) Now, if $X$ is just some arbitrary topological space, is the last formula true? My guess is that it is not, but I don't know right now what a counterexample might look like. Besides, can we show that one quantity is always bounded by the other one? Or are there examples of both types of behaviors?

It seems one cannot obtain a bound: let $T\subset X$ be an irreducible component passing through $x$, and let $U\subset X$ be an open neighborhood of $x$. Then $\dim T$ and $\dim U$ are both greater or equal that $\dim T\cap U$, but I don't see a way of comparing the values $\dim T$ with $\dim U$.

Elías Guisado Villalgordo
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  • Now that Aphelli has included counterexamples for both types of behaviors ('$>$' and '$<$'), I wonder: is there some general topological property that we can impose on a topological space so that the equality holds? (Possibly shared by $X=$ the underlying space of a scheme locally of finite type over a field.) I don't know if one could find out about this; such an $X$ has very specific topological restrictions. – Elías Guisado Villalgordo Aug 30 '23 at 14:25

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The formula does not, indeed, hold in the general case. Consider the space $X=\{a,b,c\}$ with the open subsets $\emptyset, X, \{a\}, \{a,b\}$.

Then $\dim_b{X}=1$. But $X$ is irreducible of dimension $2$, so the RHS of the formula is $2$.

To find a situation where $LHS > RHS$, consider a topological space $X=L \cup \bigcup_{n \in \mathbb{Z}}{P_n}$, where

  1. $L$ is closed, and equal to $\mathbb{C}$ with the Zariski topology,
  2. every $P_n$ is closed and isomorphic to $\mathbb{C}^2$ with the Zariski topology,
  3. a subspace $F \subset X$ is closed iff its intersection with $L$ (resp. $P_n$ for every $n$) ia closed in $L$ (resp. $P_n$),
  4. $P_n \cap L$ is the unique point $n \in L$,
  5. if $n \neq m$, $P_n$ and $P_m$ are disjoint.

Take some $x \in L \backslash \mathbb{Z}$.

Every open subset $U$ containing $x$ contains a cofinite subset of $L$, and in particular contains some $n \in L \cap \mathbb{Z}$. In particular, $U \cap P_n$ is a nonempty Zariski-open subset of $P_n$, so $\dim{U} \geq \dim{U \cap P_n} \geq 2$.

On the other hand, the only irreducible component going through $x$ is $L$, which has dimension $1$ (this isn’t entirely obvious, but I’d be very surprised if it was false). QED.

Aphelli
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  • Thanks! Three comments: (1) I think in the second line you mean "then $\dim_a{X}=0$. But $X$ is irreducible of dimension $1$, so the RHS of the formula is $1$" (2) when you write "$\dim{U} \geq \dim{U \cap P_n}$", you're implicitly using that $P_n$ is an irreducible component of $X$, right? (here's how one shows it: if $Z\subset X$ is closed irreducible containing $P_n$, then $Z=[Z\setminus(P_n\setminus{n})]\cup P_n$ is a union of two closed subsets, whence $Z=P_n$) (3) $L$ is an infinite topological space with the cofinite topology, so its irreducible subsets are $L$ and the singletons. – Elías Guisado Villalgordo Aug 27 '23 at 15:38
  • And a question: (Q) why is an open subset of $\mathbb{Q}^2$ with the Zariski topology of dimension greater or equal than $2$? (I wonder from "$\dim U\cap P_n\geq 2$" that you wrote.) Since the distinguished open sets are a basis, it suffices to show $\dim D(f)\geq 2$ for $f\in\mathbb{Q}[x,y]$, but why is so? – Elías Guisado Villalgordo Aug 27 '23 at 15:46
  • I guess one can just substitute $\mathbb{Q}$ by $\mathbb{C}$ in all of the construction and then one concludes $\dim U\cap P_n=2$ using 0A21(3) (we are allowed to use scheme theory for this reason) – Elías Guisado Villalgordo Aug 27 '23 at 16:45
  • You’re right, we should take an algebraically closed field. I edited. – Aphelli Aug 27 '23 at 19:32
  • In the beginning of my answer, I really mean $\dim_b X$ (and $X$ really has $2$). But you’re right that the spectrum of a DVR and its generic point would also work. – Aphelli Aug 27 '23 at 19:54