Let $K$ be a field and $X$ be a proper irreducible $K$-scheme. Show that the image of any $K$-morphism $X \rightarrow Y$ into an affine $K$-scheme $Y$ of finite type consists of a single point.
I came across this exercise in my reading, and was wondering how you might prove it. Any help would be appreciated!
Another argument:
The question of whether the image of $X$ is a point can be detected on the level of points sets, so it is no loss of generality to replace $X$ and $Y$ by their underlying reduced subschemes, and so suppose that they are reduced.
If $Y =$ Spec $A$, then giving $X \to Y$ is the same as giving $A \to \Gamma(X,\mathcal O_X)$. Now $X$ is complete, and hence $\Gamma(X,\mathcal O_X)$ is finite-dimensional over $K$. Indeed, since $X$ is also reduced, it is a reduced finite dimensional $K$-algebra. Since $X$ is furthermore irreducible, it must be a field. (A reduced fin. dim'l. $K$-alg. is a product of fields, but the product decomposition induces a corresponding decomposition of $X$ into connected components. Since $X$ is irred. it is connected, and so there can only be one such component.)
Thus the image of $A$ in $\Gamma(X,\mathcal O_X)$ is again a field, say $k$, and so the morphism $X \to Y$ factors as $X \to $ Spec $k \to Y,$ and hence has as image the single closed point Spec $k$ of $Y$.
The set-theoretic image of the $K$-morphism $f:X\rightarrow Y$ is irreducible. Since $f$ is quasi-compact, the underlying topological space of the scheme-theoretic image of $f$ is equal to the closure of the set-theoretic image, which will again be irreducible. Since closed subschemes of affine finite type $K$-schemes are also affine and of finite type over $K$, replacing $Y$ by the scheme-theoretic image of $f$, we may assume $f:X\rightarrow Y$ is surjective. Since $Y$ is separated and of finite type over $K$, properness of $X$ implies that in fact $Y$ is also proper over $S$ (see part (f) of Proposition 3.16 on page 104 of Liu's book). An affine $K$-scheme which is also proper over $K$ is finite over $K$. But then $Y$ is the spectrum of a finite $K$-algebra, which means its underlying topological space is discrete. Since we know that $Y$ is irreducible, $Y$ must be a single point.
