As the title says, I'm looking for an affine scheme of dimension zero, but with infinitely many points. At first I doubted that something like this could exist, and I still can't think of an example, but I'm beginning to think there must be one.
Such a scheme corresponds to a zero-dimensional ring $A$ of Krull dimension $0$, i.e. the only 'chains' of prime ideals in $A$ have to be the prime ideals themselves. If $k$ is a field, I could take $n$-fold product $k^n=k\times...\times k$. The prime ideals in $k^n$ are exactly those of the form $k\times...\times\langle0\rangle\times...\times k$, with the $\langle0\rangle$ at any position. Hence in $k^n$ there are no chains of prime ideals of length greater than $0$. But of course we only have finitely many (namely, $n$) points. Is it possible to carry this construction over to an infinite product of copies of $k$?
Take for example $k^\mathbb{N}$, which is a countable product of copies of $k$. At first I thought this was just $k[[x]]$, but the ring structure is not the same. Wikipedia states that in such an infinite product there exist ideals which are not of the type $\prod_{n\in\mathbb{N}}I_n$. An example is the ideal $I$ of elements of $k^\mathbb{N}$ with only finitely many non-zero components.
Although having an infinite indexing set, products of ideals of $k$ that contain only one factor $\langle 0\rangle$ are still prime, so we have infinitely many points in $\operatorname{Spec}(k^\mathbb{N})$. But what about the dimension? There exist other prime ideals, so I'm not sure how to go on about that. Does $k^\mathbb{N}$ work at all, or are there other - maybe simpler - examples? Or even none?
EDIT: In case of an infinite product of rings $A_i$, does the spectrum of $\prod_i A_i$ still correspond to some disjoint union of schemes? In my case, geometrically thought: by taking one more copy of $k$ into the product, we are adding a new point to our affine scheme. So from my intuition, the construction of $k^\mathbb{N}$ as an affine scheme with inf. many points and dimension $0$ should be working.
The rest of the exercise (Find examples or prove there are none. A short check if what I write is correct would be very kind):
- An affine scheme $X$ with $\dim(X)=1$ and exactly one point.
- An affine scheme $X$ with $\dim(X)=1$ and exactly two points.
As for 1: This should not be possible, since exactly one point means there is just one prime ideal, so the dimension of $X$ will be $0$.
2: $k[[x]]$ should work here, as well as any discrete valuation ring (not a field).
Thanks for reading this wall of text and for your help!