I was asked to show that if $X$ is a projective variety, $Y$ an affine variety and $\varphi:X\rightarrow Y$ is a morphism, then $\varphi(X)$ is finite. I think that I have to use this fact:
- Let $X$ be a connected projective variety and let $f: X\rightarrow \mathbb{A}^1_{\mathbb{C}}$ be a global regular map, then $f$ is constant.
This fact is proved by composing $f$ with the morphism between $\mathbb{A}^1_{\mathbb{C}}$ and $\mathbb{P}^1_{\mathbb{C}}$ that takes $z$ to $[1:z]$, and seeing that since morphisms are closed, $X$ is connected (so the image) and $[0:1]$ isn't in the image of the composition, then the image must be a unique element.
I tried to think in a way to do a similar proof but my problems are:
-
The affine variety is in $\mathbb{A}^n_{\mathbb{C}}$, maybe I have to define more then one morphism to compose ($z\mapsto [1:z_1:\dots: z_n], \dots , z\mapsto [z_1:\dots: z_n:1]$) but I'm not sure.
-
The fact is for a global regular function, now I have a morphism, I'm not seeing hot to change the arguments to make this work.
Best Answer
You can do this by applying the result you mention after doing a little prep work.
Step 1: A projective variety has finitely many connected components, so it suffices to prove the statement when $X$ is connected.
Step 2: We may assume $Y$ is $\Bbb A^n$. The statement that $Y$ is affine means that there is a closed immersion $i:Y\to \Bbb A^n$ for some $n$. We may then consider the composite $i\circ f:X\to \Bbb A^n$, and if we can show that this has finite image, we're done.
Step 3: A map $X\to \Bbb A^n$ is an $n$-tuple of regular functions. By the statement that any global regular function on a connected projective variety is constant, each coordinate must be constant, so the map $X\to\Bbb A^n$ must also be constant.