I'm looking for a proof/counterexample of the following fact:
Theorem Let $X \subseteq k^n$ and $Y \subseteq k^n$ be algebraic varieties over a field $k$ and let $\phi$ be a morphism from $X$ to $Y$. If the induced morphism on the Coordinate Rings is integral, then the fibers of $\phi$ are finite.
Please, help me to prove or disprove this fact or give me some references.
Thank you!
Best Answer
Let $A$ be the coordinate ring of $X$ and let $B$ be the coordinate ring of $Y$. Since $\phi : X \to Y$ is a morphism, we get a homomorphism $\phi^* : B \to A$. To show that the fibres of $\phi : X \to Y$ are finite, it is enough to show that $A$ is a finite $B$-module (via $\phi^* : B \to A$), because finiteness is preserved under base change.
So suppose $A$ is integral over $B$. We know $A$ is finitely generated as a $k$-algebra, so it is finitely generated as a $B$-algebra a fortiori, say by $a_1, \ldots, a_r$. But each $a_i$ is integral over $B$, so there exist a natural number $N$ such that $a_1, \ldots, a_1^N, \ldots, a_r, \ldots a_r^N$ generate $A$ as a $B$-module. Thus $A$ is indeed a finite $B$-module.