# Morphisms from a proper scheme to the affine line over a field must be constant.

Let $k$ be a field. Let $X$ be a (non empty) connected proper $k$-scheme. I would like to prove the maximum principle, that is for any $k$-morphism $\varphi \colon X \to \mathbb A_k^1$, the image of $\varphi$ is a closed point.

From a previous result, as $\mathbb A_k^1$ is a separated $k$-scheme, I already know that $\varphi(X)$ is closed in the affine line. It remains to show that $\varphi(X)$ is a single point. I have no clue…

I tried to exploit the fact that a morphism $\varphi \colon X \to \mathbb A_k^1$ is given by a morphism $k[T] \to \mathscr O_X(X)$ (via the left adjoint of the inclusion of affine schemes into the locally ringed spaces) and it lead me to exhibit $\varphi$ as follow : there is an global section $f \in \mathscr O_X(X)$ such that for all $x \in X$, $$\varphi(x) = \ker (k[T] \to \kappa(x) \colon Q \mapsto Q(f)) .$$ But I don't see why this prime ideal of $k[T]$ wouldn't change with respect to $x$.

First we claim that the image of $\varphi$ cannot be all of $\Bbb{A}^1$. If it were all of $\Bbb{A}^1$, by composing with $\Bbb{A}^1 \hookrightarrow \Bbb{P}^1$ we get a dominant map $X \to \Bbb{P}^1$ that is furthermore proper and thus has closed image. But now this means the map $X \to \Bbb{P}^1$ is surjective, a contradiction.

So we know the image of $\varphi$ is a proper closed subset of $\Bbb{A}^1$. It is now enough to observe that the image of a connected set under a continuous function is also connected. For then the only connected proper closed subset of $\Bbb{A}^1_k$ is a one point set.

Added: Why is $X \to \Bbb{P}^1$ proper?

Well this comes from the famous "Property $\mathscr{P}$" exercise:

Let $\mathscr{P}$ be a property of morphisms that is stable under composition and base extension. Then suppose $f : X\to Z$ which factors as $X \stackrel{g}{\to} Y \stackrel{h}{\to} Z$. If $f$ and the diagonal $\Delta{h}$ have property $\mathscr{P}$, then so does $g$.

• I think you meant $\mathbb A_k^1$ instead of $k$. Just to be sure I'm getting it right : a closed subset of $\mathbb A_k^1$ is (topologically) $\mathrm{Spec}\,(k[T]/I)$ for some ideal $I$ of $k[T]$ ; but a scheme (in fact, a locally ringed space) is connected if and only if its global section ring does not have non trivial idempotent, which is in turn equivalent to be unwritable as a product of two non trivial rings ; in conclusion, a closed subset $V(I=(Q)) \simeq \mathrm{Spec}\, (k[T]/(Q))$ is connected iff $Q$ irreducible or null (by the Chinese theorem). Is that it ? – Pece Jan 4 '14 at 14:21
• … And if it is, why can't $\varphi(X)$ be the connected closed subset $\mathbb A_k^1$ (that is $Q=0$) ? – Pece Jan 4 '14 at 14:31
• @Pece For your first statement, you're really overcomplicating it: All it is saying is that a polynomial of degree $n$ can have at most $n$ number of roots! – user38268 Jan 4 '14 at 14:53
• @Pece For your second comment- I overlooked that part, and have now edited my answer. – user38268 Jan 4 '14 at 15:03
• I get the first part, embedding $\mathbb A^1$ into $\mathbb P^1$. For the second part, I don't get why you mean by "All it is saying etc." : from where do you see (without my overcomplicating explanation) that $V((Q))$ is connected iff $Q$ is irreducible ? – Pece Jan 4 '14 at 15:10