Algebraic Geometry – Why Dimension of Generic Fiber Equals Transcendence Degree of Function Fields?

algebraic-geometry

I am reading the Gortz's Algebraic Geometry book, p.466, statement of Lemma 14.106 and and some question arises :

Lemma 14.106. (1) Let $X$, $Y$ be irreducible locally noetherian schemes, let $f:X\to Y$ be a dominant morphism locally of finite type, and denote by $\xi$ and $\eta$, resp., the generic points of $X$ and $Y$. Let $e:=\dim f^{-1}(\eta)=\operatorname{trdeg}_{\kappa(\eta)}\kappa(\xi)$ be the dimension of the generic fiber. Let $x\in X$ and $y=f(x)$. Then
$$ e+ \dim\mathcal{O}_{Y,y} \ge \operatorname{trdeg}_{\kappa(y)}\kappa(x) +\dim \mathcal{O}_{X,x} \tag{14.26.1}$$

Q. Why can we write $\dim f^{-1}(\eta)=\operatorname{trdeg}_{\kappa(\eta)}\kappa(\xi)$ ?

Perhaps, can we try to use next theorem ( Gortz's book, p.126, Theorem 5.22- (1) ) :

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?

In applying the Theorem 5.22-(1), I think that there are several technical diffiulties

Q.1 If we use the theorem 5.22 'directly', then the generic fiber $f^{-1}(\eta)$ is irreducible? I found an assoicated theorems ( Gortz's book, Exercise 6.15. )

Exercise 6.15. Let $Y$ be an integral scheme with generic point $\eta$ and let $f:X\to Y$ be a morphism. Let $X$ be reduced ( resp. irreducible, resp. normal, resp. regular ). Show that the generic fiber $X_{\eta}$ has the same property.

Can we try to use this exercise? A technical subtlety is that in our situation ( Lemma 14.106 ) , $Y$ may not be intergral scheme. By introducing its underlying reduced subscheme $Y_{\operatorname{red}}$ of a scheme $Y$, we can deduce that $f^{-1}(\eta)$ is irreducible ?

Q.2 In the above lemma 14.106, since $f$ is dominant, $\xi \in f^{-1}(\eta)$. Then

Q. 2-1) $\xi$ is generic point of $f^{-1}(\eta)$ ( if $f^{-1}(\eta)$ is irreducible )? I think that this is true as Closure of a subset of a subspace of a topological space.

Q. 2-2) If Q. 2-1) is true, then the residue field of $\xi$ 'in $f^{-1}(\eta)$' is same as the residue field $\kappa(\xi)$ 'in $X$' ?

If these questions Q.1, Q.2 are true, then we may apply the Theorem 5.22 and get our disired result.

Can anyone help?

Best Answer

Since every adjective here is preserved by passing to an open subscheme, we can assume that $X=\operatorname{Spec} A$ and $Y=\operatorname{Spec} B$ (also, since the inclusion $\operatorname{Spec}\kappa(\eta)\to Y$ factors through any non-empty open immersion, the fiber $f^{-1}(\eta)$ is not changed). $f$ is now necessarily finite type because it's affine. We can also assume that $A$ and $B$ are integral domains since reduction preserves the underlying topological space of the fibers and the residue field at the generic point. The fact that $f$ is dominant then implies that $B$ is a subring of $A$. Then $f^{-1}(\eta)=\operatorname{Spec}(A\otimes_B K(B))$ is the spectrum of a localization of $A$ (which is an integral domain) and thus irreducible. The residue field at the generic point of $f^{-1}(\eta)$ is $K(A\otimes_B K(B))=K(A)=\kappa(\xi)$. $f^{-1}(\eta)$ is now an irreducible finite type scheme over $K(B)=\kappa(\eta)$ and thus $\dim f^{-1}(\eta)=\operatorname{trdeg}_{\kappa(\eta)} \kappa(\xi)$.

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