6
$\begingroup$

Just wondering about conventions dealing with the zero ring and the zero scheme.

  1. Does the category of schemes have an inital object?
  2. Is the zero ring considered local?
  3. For the purposes of scheme theory, is a map of sheaves that induces on stalks a map of the form $\mathcal{O}_{X,P}\to 0$ considered a "local" homomorphism on stalks?

https://en.wikipedia.org/wiki/Zero_ring Wikipedia says that the zero ring is not local.

I am wondering how to square this with certain conventions in scheme theory. If $0$ is the zero ring, then conventionally (say in Hartshorne Chapter II, exercise 2.6) the category of schemes has $Spec(0)$ for an initial object; whose underlying space is $\emptyset$ and whose associated sheaf is the constant sheaf at zero. The direct image of this sheaf along the canonical map $\emptyset \to X$ would be, again, the constant zero sheaf, so the canonical natural transformation $\mathcal{O_X}\to 0$ would send every stalk to zero. It seems like this shouldn't count as a map of locally ringed spaces.

$\endgroup$
1
  • 3
    $\begingroup$ As Hartshorne points out, the convention is that the spectrum of the zero ring is the initial object and the underlying set is empty. This squares well with the Wikipedia claim that the zero ring is not local. Indeed if it was local then it would be a one point set, since it would have a unique prime (maximal in fact) ideal. $\endgroup$
    – Luke
    Sep 20, 2020 at 2:47

3 Answers 3

10
$\begingroup$

The empty scheme is initial in the category of schemes, and the zero ring is not a local ring, since it does not have a unique maximal ideal (it does not have any maximal ideal!). There is no special convention needed here--this all just follows from the general definitions.

In particular, there is no issue with what the unique map out of the empty scheme does on stalks. If $X$ and $Y$ are locally ringed spaces, then a morphism $X\to Y$ is a continuous map $f:X\to Y$ together with a morphism of sheaves of rings $\mathcal{O}_Y\to f_*\mathcal{O}_X$ such that for each $x\in X$ the induced map on stalks $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is a local homomorphism. When $X$ is empty, there are no points $x\in X$ at which to check this condition, and so it holds vacuously.

$\endgroup$
1
  • $\begingroup$ Thank you. My problem was that I neglected the bolded part. $\endgroup$ Sep 20, 2020 at 11:23
8
$\begingroup$
  1. The empty scheme is the initial object.

  2. The zero ring is not a local ring.

  3. Conventionally local ring homomorphisms are between local rings, but we may extend the definition to general rings by defining “local” to mean that an element becomes invertible in the codomain if and only if it is invertible in the domain. Under this definition a ring homomorphism to the zero ring is local if and only if the domain is the zero ring. Anyway this is irrelevant to your question about the empty scheme: it has no points, so its structure sheaf has no stalks, so the condition is vacuous.

$\endgroup$
7
$\begingroup$

The category of schemes has an initial object, which is affine, given by the spectrum of the zero ring, which empty. The zero ring is not a local ring; a local ring has to have a unique maximal ideal, and the zero ring doesn’t have any (it’s the only ring with this property), because it is not a field.

If you take out the empty scheme then the resulting category will fail to have fiber products.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .