Let $f: X \rightarrow Y$ be a morphism of irreducible projective varieties, that is both finite and surjective.
Does this mean that it is flat?
I have tried the following:
By finiteness, the map is locally $\text{Spec}(B) \rightarrow \text{Spec}(A)$, i.e. $A \rightarrow B$ in rings, where $B$ is a finitely generated $A$-module. By surjectiveness, A goes injectively into $B$. Since we are dealing with varieites, $A$ and $B$ are finitely generated $k$-algebras with no nilpotents. In fact since the varieties are irreducible, they have no zero divisors.
Since a module is flat over a ring if and if for all prime ideals in the ring, the localization of the module is flat over the localized ring, we would be done if we could show that $B$ is flat over $A$.
This is where my knowledge gets too sketchy. We cannot assume $A$ to be a PID right? Is it true that we exactly need to show that $B$ is an acyclic object in $A$-Mod for $\text{Tor}$? Finally, can someone tell me how to show that $B$ is flat over $A$, or tell me that this is wrong?
Thanks!