Morphism between varieties

769 Views Asked by At

Let $V$ and $W$ be two (irreducible) varieties over an algebraic closed field $k$. Then there is a definition of what a morphism of varieties $f:V\rightarrow W$ is. On the other hand we can see $V$ and $W$ as locally ringed spaces with the sheaves of regular functions $\mathcal{O}_V$ and $\mathcal{O}_W$ respectively. There is also a definition of what a morphism of locally ringed spaces $(f,f^\sharp):(V,\mathcal{O}_V)\rightarrow (W,\mathcal{O}_W)$ is (Hartshorne, Chap 2, Sec 2).

My question is: is it the same to give a morphism of varieties $f:V\rightarrow W$ and to give a morphism of locally ringed spaces $(f,f^\sharp):(V,\mathcal{O}_V)\rightarrow (W,\mathcal{O}_W)$?

1

There are 1 best solutions below

3
On BEST ANSWER

Yes, this is the fully-faithfulness part of the well-known equivalence of categories between classical varieties and varieties in scheme theory. See Hartshorne II.2.6 or Görtz-Wedhorn 3.37.

In my opinion this equivalence of categories is best seen as an adjoint equivalence (instead of a functor which is fully-faithful and essentially surjective). That is, for every variety $X$ we construct a corresponding scheme $\tilde{X}$ by gluing and conversely for every scheme $Y$ of finite type over $k$ we endow the rational points $Y(k)$ with the structure of a variety. These are functors, and there are natural morphisms $Y \to \widetilde{Y(k)}$ (unit) and $\tilde{X}(k) \to X$ (counit) satisfying the zig-zag identities, i.e. we have an adjunction. Then one proves that unit and counit are isomorphisms, which can be checked locally and thereby reduced to affine varieties resp. affine schemes.