# constant morphisms from projective, integral schemes over k

Let $$X$$ be a projective, irreducible, reduced scheme over $$k$$ and $$Y$$ be an affine $$k$$-scheme, where $$k$$ is algebraically closed. Prove that every morphism $$f : X → Y$$ is constant.

I know that for a general scheme $$X$$ which is irreducible as well as reduced (and $$Y$$ is just given to be a scheme i.e. no extra conditions imposed), combined with the fact that $$f$$ restricted to every affine open subset constant would imply that $$f$$ is constant . So if I can basically show that restriction of $$f$$ to every affine open subset is constant, we're done.

Now if in addition $$Y$$ is assumed to be an affine scheme, say $$\operatorname{Spec}(R_1)$$ then take any affine open subset say $$\operatorname{Spec}(R_2) \subset X$$ , then the homomorphism induced on the rings $$R_2 \to R_1$$ associated from the restriction of the morphism to the affine open subset, goes in the global section of the structure sheaf.

How to maybe combine these with the $$k-scheme$$ structure and also how to use the fact that our $$X$$ is also projective and a scheme over $$k$$.

Thanks for help.

• 1. If you use \operatorname{Spec} to format $\operatorname{Spec}$, your text will look nicer. I've made this adjustment for you. 2. Hint: have you heard that a map to an affine scheme is determined by the map on global sections? This will probably be more effective than your current strategy for this problem. Jun 22 '20 at 22:44
• @KReiser Actually I haven't. Can you kindly elaborate your thought as an answer. Jun 22 '20 at 22:48

First, we note that $$\mathcal{O}_X(X)$$ is a field since $$X$$ is an integral projective scheme over a field. Next, for any locally ringed space $$X$$ and any affine scheme $$Y$$, there is a bijection between morphisms $$X\to Y$$ and ring homomorphisms $$\mathcal{O}_Y(Y)\to \mathcal{O}_X(X)$$ (see for instance Stacks 01I1, or EGA III Err 1 Prop 1.8.1 where it is attributed to Tate, or Hartshorne exercise II.2.4). The interesting portion of this bijection is that given a map on global sections $$f$$, we send a point $$x\in X$$ to the point $$y\in Y$$ corresponding to the preimage of $$\mathfrak{m}_x\subset \mathcal{O}_{X,x}$$ under the composite map $$\mathcal{O}_Y(Y)\stackrel{f}{\to}\mathcal{O}_X(X)\to\mathcal{O}_{X,x}.$$
So the morphisms $$X\to Y$$ are in bijection with maps from $$\mathcal{O}_Y(Y)$$ to a field. But such a map is given by a maximal ideal of $$\mathcal{O}_Y(Y)$$, and tracing through the bijection above, we see that this means that all points $$x$$ map to this maximal ideal.
To give an intuitive complement to the other answer, one can think of global sections of the structure sheaf of $$(X,O_X)$$ as maps to $$X\rightarrow \mathbb{A}^1$$, and affine schemes have heaps of such maps, while projective schemes have very few. One can then prove the result by noting that if the map $$X\rightarrow Y$$ wasn't constant, we could find a map from $$Y$$ to $$\mathbb{A}^1$$ that witnesses this ("functions on affine schemes seperate points"). Justifying this will boild down to unravelling the definitions of what it means to be a constant map, and using that the global sections of $$O_X$$ are just $$k$$, but I find it easier to do so with this geometric picture in mind.