A question in Mumford Redbook P91

Question

Suppose $K_0$ is a field isomorphic to the function field of prevariety$X,Y$.

Choose $k-$isomorphism $\alpha:K(X)\to K_0; \beta: K(Y)\to K_0$,

we get $A: \operatorname{Spec} K_0\to X, B:\operatorname {Spec}K_0 \to Y$,

so we get $(A,B): \operatorname{Spec}K_0\to X\times_kY$, the image of the closed point is $t$. Take its closure we get irreducible subprevariety $T=\bar{\{t\}}$ .

How do we show the function field $K(T)$ of $T$ is isomorphic to $K_0$ via $(A,B)^*$ ?

Consider the simple case where $X=\operatorname{Spec}k[S],Y=\operatorname{Spec}k[T], K_0=K(U)$, (the rational function field of one variable)$\alpha: S\to U, \beta: T\to \frac{1}{U}$, then which point in $\mathbb{A}^2$ is mapped to? is it the generic point of curve $ST=1$? How to make this clear?

(It is in Mumford's redbook II.3 P91 )

Answer 1

a) By the categorical definition of product the composition $\operatorname {Spec} K_0 \to T \to X$ is the morphism $A: \operatorname {Spec} K_0 \to X$, so that we have morphisms of $k$-extensions $k(X)\to k(T)\to K_0$ whose composition is the isomorphism $\alpha: k(X)\to K_0$.
This implies that $ k(T)\to K_0$ is surjective and, as it is injective (like all morphisms between fields!), it is an isomorphism, which answers your question.

b) You are absolutely correct about your beautiful example:
$\operatorname {Spec}(K_0)$ maps to the point $p\in \mathbb A^2_k=\operatorname {Spec}(k[S,T])$ given by the prime ideal $\mathfrak p=(ST-1)$ whose residue field is $\kappa(p)=k(S)$ ,
the dual map being $\kappa(p)=k(S)\to k(U)=K_0: S\mapsto U$
The point $p$ then projects to the generic points $x,y$ of both copies of the affine line and for example the morphisms of $k$-extensions $\kappa (y)=k(T)\to \kappa(p)=k(S)\to K_0$ are characterized by $T\mapsto S^{-1} \mapsto U^{-1}$