What are the closed sets on $\mathbb{Z}$, $\mathbb{R}[x]$, $\mathbb{Z}[x]$? I was given this question, but it seems trivial, because aren't the closed sets only the affine varieties? So for each is just the irreducible varieties?
So I understand better, could you also give me an example of a set in one of those that is not closed? Would one be a line without the origin?
Edit: this is with regards to the spectrum of a ring R, defined to be V(S)=$\{P \in SpecR | S \subseteq P\}$ where P is a prime ideal. Then the collection $\{V(S)|S\subseteq R \}$ determines a topology on SpecR, the zariski topology. (If someone could explain why this is true, that would be great too!)
Best Answer
I) The irreducible closed subsets of $\mathbb A^1_\mathbb Z=\operatorname {Spec}\mathbb{Z}[X]$ are exactly the following, listed by dimension:
Dimension two
$\mathbb A^1_\mathbb Z=V(0)$
Dimension one
$\bullet$ The curves $V(f) $ where $f\in \mathbb Z[X]$ is an irreducible polynomial
$\bullet \bullet$ The curves $\mathbb A^1_{\mathbb F_p}=V(p)$ here $p$ is a prime integer.
Dimension zero
The closed points $V(p,f(X))$ where $f(X)\in \mathbb Z[X]$ is a polynomial whose reduction $\overline {f(X)}\in \mathbb F_p[X]$ modulo the prime $p$ is irreducible.
II) An arbitrary closed subset of $\mathbb A^1_\mathbb Z$ is a finite union of some of the above irreducible closed subsets of $\mathbb A^1_\mathbb Z$, where mixing subsets of different dimensions is of course allowed.
[The empty set is thus closed : it is the union or the empty set of irreducible closed subsets.
If you are allergic to such set-theoretical sophistry, just explicitly add $ \emptyset$ to the above description]