Schemes 10: Morphisms of affine schemes

This lecture is part of an online algebraic geometry course on schemes, based on chapter II of "Algebraic geometry" by Hartshorne.
We try to define morphisms of schemes. The obvious definition as morphisms of ringed spaces fails as we show in an example. Instead we have to use the more subtle concept of a morphism of locally ringed spaces. This can then be used to show that morphisms of rings correspond exactly to morphisms of affine schemes. We conclude by using this to show that the plane minus the origin is not an affine scheme.

[autumnsthree]

I figure out. The comment below is nonsense. Since we only require the hom on the stalk is local hom, so Q -> Q is local hom but Q -> Z(p) is not local hom, which makes second hom not morphism of locally ringed space.

15:00 - 16:00 is a bit confusing. Since (p) is actually max ideal of Z(p) and (0) is max ideal of Q, so why 2nd hom of ringed space which map max ideal to max ideal is non-local and bad morphism?

[hausdorffm]

For any homomorphism of rings R -> S, there is a induced map Spec(S) -> Spec(R).
In general, for a map Spec(S) -> Spec(R), we cannot find a homomorphism of rings R ->S.

Example 1, R = Z = integers and S = Q = rational numbers.
In this case, there is only one homomorphism R -> S.
On the other hand, we can take many maps Spect(Q) -> Spec(Z).
For example, Spec(Q) -> Spec(Z); (0) -> (p) for any prime p of Z and for p = 0.
Namely,


Spect(Q) -> Spec(Z); (0) -> (0) (Induced map on Stalks of (0) are local homorphism)

Spect(Q) -> Spec(Z); (0) -> (2) (Stalks of (0) has maximal ideal Q, but a maximal ideal of stalk of (2) is not Q)

Spect(Q) -> Spec(Z); (0) -> (3)

Spect(Q) -> Spec(Z); (0) -> (5)

Spect(Q) -> Spec(Z); (0) -> (7)

Spect(Q) -> Spec(Z); (0) -> (11)

Spect(Q) -> Spec(Z); (0) -> (13)
:
:


So, the map Spect(Q) -> Spec(Z); (0) -> (p) can be induced by homomorphism of rings if p = 0.


To exclude the maps Spect(Q) -> Spec(Z) which is not induced from homomorphism of rings Q -> Z, we use the notion of LOCAL HOMOMORPHISM. Restricting maps of affine schemes on notion of local homomorphism, we can always say that any map of affine schemes are induced from map of rings.


Example 2. Let Z be integers and Q be rationals. Let R be the localization of Z at some fixed prime ideal (p), i.e., R = Z_(p).
Let S be a ring of rationals, i.e., S = Q.

Then, Spec(R) = {(0), (p)} and Spec(S) = { (0) }.

So, there are two maps:

Spec S -> Spec R; (0) -> (0),

and


Spec S -> Spec R; (0) -> (p).

On the other hand, only one possible ring homomorphism R -> S.

To make Spec() to be good as a functor, we want to exclude
the maps of affine schemes which is not induced from ring homomorphism. To do so, we use the notion of local homomorphism.


In case of non affine schemes X, Y,
even if rings of global sections of X and Y are isomorphic,
it does not mean that schemes are also isomorphic.
As a counter example, see at 26:10.
The scheme of punctured plane and plane has same rings of global sections of sheaf.
However, as schemes they are not isomorphic.

[MudithaMaths]

Why did the professor say that the definition of a "natural transformation" is hairy?