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!
No. Let $C$ be a projective curve which has only one singularity, a node. Then the normalization map $\tilde{C} \rightarrow C$ is not flat, but it is finite and surjective.
Surjectivity is clear. Finiteness is by the finiteness of normalization.
If the map is flat, then as it is flat and finite, all fibers should have the same size, which is false here.
I believe that more generally, normalization is typically not flat; but I don't know what hypothesis you need to say this. Perhaps it's true whenever the variety isn't already normal.
EDIT: I found https://mathoverflow.net/questions/64776/flatness-of-normalization, so apparently this is true whenever the variety isn't already normal.