Spec of $k(X)$ is the fibre of $f$ over the generic point of $Y$ given $f: X \to Y$ I am trying to understand a part of proof of Corollary 2  in Section III. 5
in Mumford's Red book.
Suppose I have an etale dominant morphism $f: X \to Y$ where $X$ and $Y$ are separated irreducible reduced scheme of finite type over an algebraically closed field.
He states that the Spec of $k(X)$, the function field of $X$, is the fibre of $f$ over the generic point of $Y$.
How can I prove the statement?
I think I have covered all the assumptions in the proof, but the example provided in the comment looks like a counterexample... I would appreciate any clarification or explanation on this. Thank you.
ps In this section Mumford defines etale for $f$ of finite type.
 A: Let $\eta=\operatorname{Spec}k(Y)$ be the generic point of $Y$, then if we basechange $X \to Y$ via $\eta \to Y$ then we get $X_{\eta} \to \operatorname{Spec}k(Y)$, which will be étale. In other words, we have a finite étale algebra $A$ over $k(Y)$ of degree $d$ with spectrum equal to $X_{\eta}$, and we want to show that $A \simeq k(X)$ as $k(Y)$ algebras.
But the universal property of the fiber product gives us a map $A \to k(X)$ of $k(Y)$-algebras, and to prove that this is an isomorphism it suffices to prove that it is injective [because it is a morphism of finite dimensional $k(Y)$ vector spaces of the same rank].
To show that it is injective, it suffices to show that $A$ is a field. First of all we know that $X_{\eta}$ is homeomorphic to $f^{-1}(X_{\eta})$ [c.f. https://math.stackexchange.com/q/2439943/45878], and so that $\operatorname{Spec} A$ has precisely one point. Indeed, the only point of $X$ that maps to the generic point of $Y$ is the generic point of $X$, because $X$ and $Y$ are equidimensional. It follows that $A$ is the spectrum of a finite field extension of $k(Y)$ (an étale algebra over a field is always a product of finite field extensions).
