Let $f:X \to Y$ be a morphism of projective varieties (integral schemes of finite type) over a field $k$. If $f$ is dominant, then $f$ induces a field extension $K(Y)\to K(X)$ of function fields. We call $f$ finite whenever this extension is. The degree of f is then defined to be the dimension of $K(X)$ as a vector space over $K(Y)$.
My question is this: what are the possible implications between $f$ having finite degree and $f$ being a finite morphism? I think that $f$ having finite degree would imply that all its fibers are finite, which in turn implies that $f$ is finite, but I'm not sure how to go about this. For the other direction, I have no idea where to start.
Finite implies finite degree, but finite degree does not imply finite (nor quasi-finite).
Finite implies finite degree: let $\eta_X$, $\eta_Y$ be the generic points of $X$ and $Y$, thought of as the spectra of their function fields. Then the fiber product of $i:\eta_Y\to Y$ and $f$ is a finite scheme over $\eta_Y$ which must be $\eta_X$ by integrality of $X$, so $K(Y)\subset K(X)$ is finite.
Finite degree does not imply finite: consider a blowup of $\Bbb P^2$ in a closed point $P$. Then this is a birational morphism (hence of degree 1) but not finite.