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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.

Johnny T.
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    Have you taken a look at the projection $\mathbb{A}^2\to \mathbb{A}^1$ ? – Captain Lama Jul 20 '20 at 13:06
  • Let me change the question – Johnny T. Jul 20 '20 at 13:56
  • The same example I gave shows that the étale condition is needed for the statement (the projection $\mathbb{A}^2\to \mathbb{A}^1$ is clearly not étale, for instance). – Captain Lama Jul 20 '20 at 15:11
  • oops yes, let me just add etale – Johnny T. Jul 20 '20 at 15:27
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    Consider a map $X \rightarrow Y$ of elliptic curves over $k$ of degree $d > 1$. It satisfies all of the given hypotheses, but the generic fiber has $d$ elements. – MarkM Jul 20 '20 at 19:56
  • I'm trying to understand a part of a proof in Mumford. I will add this in the question. – Johnny T. Jul 21 '20 at 07:44
  • Consider an etale morphism $X \to Y$ of elliptic curves of degree $d>1$, then this corresponds to an unramified extension of the function fields $k(Y) \subset k(X)$. The degree of the morphism $\operatorname{Spec} k(X) \to \operatorname{Spec} k(Y)$ will be equal to $d$, but $\operatorname{Spec} k(X)$ is irreducible. Of course if we look at a geometric generic fibre (base change to the algebraic closure of the generic point of $Y$), then the fiber will be a disjoint union of $d$ points. – Pol van Hoften Jul 21 '20 at 09:52

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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).

Pol van Hoften
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  • How do we know that $A$ and $K(X)$ have the same rank? – Johnny T. Jul 24 '20 at 13:07
  • Also what exactly is $A$ here? – Johnny T. Jul 24 '20 at 14:40
  • I have used $A$ to denote the $k(Y)$-algebra with $\operatorname{Spec} A= X_{\eta}$. We know that the ranks are the same because the degree of a finite étale morphism is constant (and it can be computed on the generic points as the degree of the field extension). – Pol van Hoften Jul 24 '20 at 17:47
  • I see, thank you. I don't quite follow how we know $X_{\eta}$ is an affine scheme... – Johnny T. Jul 24 '20 at 20:12
  • Finite morphisms are in particular affine, so if we basechange by an affine scheme (e.g. the generic point of $Y$), then we get an affine scheme over $\operatorname{Spec} k(Y)$. – Pol van Hoften Jul 25 '20 at 04:16