So I'm new to this whole scheme theory business. I'm working my way through Gortz and I produced a solution to an exercise but it seemed too easy. I'm hoping someone can either tell me that I am indeed doing things correctly, or show me in what way my proof is insufficient. This isn't homework but just my own personal reading. I know it's probably elementary, but please bare with me.
The Statement Let k and k' be fields of differing characteristic. Let X and X' be schemes over k and k' respectively, X $\ne \emptyset$. Then, there are no morphisms X $\rightarrow$ X'.
My Suggestion Suppose $f$:X$\rightarrow$X'. Let p and q be the characteristic of k and k' respectively. Then, $f^\#_x : O_{Y,f(x)} \rightarrow O_{X,x}$ is a local homomorphism of rings. Since $f^\#_x$ is local it extends to a map $O_{Y,f(x)}/M'$ to $k(O_{X,x})/M$ of the relevant fields given by taking the quotient by the relevant maximal ideals. We claim $O_{Y,f(x)}/M'$ and $O_{X,x}/M$ have characteristic p and q respectively, because they are the quotient of direct limits of rings of characteristic p and q.* But the category of fields is not connected, and has connected component the fields of various characteristic. Hence, there can be no such morphism.
- * So this is the assertion I am least sure of. Is characteristic preserved in the inductive limit? If so, what would be a good reference for me to see the proof of this, or is it obvious and I'm just not seeing it?
Lastly, I want to know if this is a general sort of argument in scheme theory. Should I try this approach more often when disproving the existence of morphisms, or is passing to the quotient of the stalk not the right idea?
Applying the global section functor $\Gamma$ to the given morphisms we obtain ring morphisms $k'\to \Gamma (X', \mathcal O_{X'}) \to \Gamma (X, \mathcal O_{X})$ and $k\to \Gamma (X, \mathcal O_{X})$.
The second morphism and the composition of the first two ones show that we have ring morphisms $k'\to \Gamma (X, \mathcal O_{X})$ and $k\to \Gamma (X, \mathcal O_{X})$.
In other words the ring $\Gamma (X, \mathcal O_{X})$ is both a $k$-algebra and a $k'$-algebra.
But a non-zero ring $A$ like here $\Gamma (X, \mathcal O_{X})$ cannot simultaneously be an algebra over two fields $k,k'$ of different characteristics because the characteristic of a non-zero algebra over a field is the same as that of the field.
This contradiction shows that no scheme-morphism $X\to X'$ exists.