Hartshorne Exercise 1.1.10: Give an example of a noetherian topological space of infinite dimensions

Question

Hartshorne Exercise 1.1.10: Give an example of a noetherian topological space of infinite dimensions.

I'm baffled by why such space can exist. Instincts told me that I shouldn't take the Spec of any Noetherian ring because then prime ideals have finite height, which makes the dimension of the Spec to be finite as well. But this still doesn't make sense for the following reason: by part $(a)$, we know that $X$ is a topological space which is covered by a family of open subsets $\{U_i\}$, then $\dim X = \sup \dim U_i$. $X$ by being a Noetherian topological space, is campact and thus we can have a finite sub-cover of the cover, and $\dim X = \max U_j$, $j \in J$ where $J$ is a finite index set. Since each $U_j$ is then finite dimensional space, we thus have $\dim X$ in finite.

Am I missing something? What would be an example?

This is true if you have an open cover that doesn't require the open set $X$. But, some spaces have no covers that don't include the whole space. — J126, May 16 '18 at 19:27
There are Noetherian rings of infinite Krull dimension. Consequently, the spectra of such rings also have infinite dimension as topological spaces. — isekaijin, Oct 03 '19 at 19:53

score 7 · Answer 1 · answered May 16 '18 at 19:29

7

An example of such a space would be the unit interval where you make the closed intervals $\left[ \frac{1}{n}, 1 \right]$ the closed sets.

answered May 16 '18 at 19:29

J126

17,707

Thank you! But is $X$ still a topological space? – nekodesu May 16 '18 at 20:53
@octoberbear The intervals I defined are closed under intersections and finite unions. Thus they define a topology. – J126 May 17 '18 at 00:54
Thanks! But if your underlying set is the unit interval, if must satisfy the topology axioms to become a topological space, which means we also need $Ø$ and $X$ to be in the topology. – nekodesu May 17 '18 at 01:45
@octoberbear Yes, but you can add those in at no extra cost. – J126 May 19 '18 at 01:23
What do you mean? – nekodesu May 19 '18 at 15:01
@octoberbear Just take the sets I described and add in both $\emptyset$ and $X$. Unions and intersections still work out. – J126 May 19 '18 at 15:03

score 7 · Answer 2 · answered May 08 '20 at 13:20

7

Similarly take $\mathbb{N}$ with the topology where the closed sets are {1,2,...,n} for any n.

answered May 08 '20 at 13:20

Pacifism

196

score 1 · Answer 3 · answered Aug 07 '21 at 03:44

prime ideals have finite height, which makes the dimension of the Spec to be finite as well

This is not true, because you could find longer and longer chains of prime ideals, each becoming stationary after finitely many steps but there is no global bound before which each and every one of them becomes stationary.

For the example see Stackexchange question

Hartshorne Exercise 1.1.10: Give an example of a noetherian topological space of infinite dimensions

3 Answers3