Can there be non open affine subset in a scheme?

Question

Now this may sound like a stupid question but I am studying schemes' theory and open affine subset of a scheme arise naturally in many statements. In many cases I wonder if the request on openness of an affine subset of a scheme in the hypothesis of a theorem is really necessary or if any affine subset of a scheme is necessarily open.

Answer 1

There are two kinds of "natural" affine subsets of (affine) schemes:

1. Localisations: This is the one you are alluding to probably. For $R$ a ring and $f\in R$, one has a natural inclusion $$Spec(R_f)\to Spec(R)$$ and this map is an embedding with image $D(f)=Spec(R)-V(f)$, thus open

2. Quotients by ideals: Take any ideal $I\subset R$, then there is again a natural inclusion $$Spec(R/I)\to Spec(R).$$ You can check that this is an embedding with image $V(I)\subset Spec(R)$, which by definition is closed.

Thus we need to find some $Spec(R)$ with $V(I)$ not open.

Take for example $R=k[x]$ for some field $k$ and $I=(x)$. Then this corresponds to the inclusion of a point into the affine line $$ *\to \mathbb{A}^1_k.$$

Answer 2

The other answers give fine examples, but I think it is worth to first point out that "an affine subset" of a scheme does not truly make sense. A subscheme can be affine or not, but a subset of a scheme does not in general have a natural structure of subscheme: it can have none, and it can have several.

It turns out that if the subset is open, then it always has a canonical structure of subscheme, so for open subsets it makes sense to ask if it is affine or not.

Now to highlight how far affine subschemes are from being open: if $X$ is an affine scheme, every closed subscheme is affine. So an affine subscheme not being open is not a weird phenomenon that you can find examples of if you look hard enough: it is completely the norm, and is a fundamental part of the theory.