Number of points in the fibre and the degree of field extension

Let $X,Y$ be varieties over $\mathbb{C}$, $k(X), K(Y)$ be function fields of $X, Y$. Suppose $\pi: X \to Y$ is a dominant, $\textit{injective}\ $ morphism, why the degree of the function field extension $[K(X) : K(Y)] =1$?

If $\phi : X \to Y$ is a finite morphism, then the fibre is finite, and by semicontinuity theorem, let $n$ be the number of the points in the generic fibre, then I feel one should similarly have $[K(X) : K(Y)] =n$. But I don't know how to show that. Any suggestions or reference on this question?

$\textbf{Edit}$: I really want the morphism $\phi$ to be a morphism between locally finite type and finite morphism. To be precise, for any affine open set $U=\rm{Spec}(B) \subset Y$, there is an affine open over of $\pi^{-1}(U)$, such that each $\rm{Spec}(A_i)$ in this cover has the property $A_i$ is a finitely generated $B-$module. I don't know the corresponding definition of this sort of morphism, or is it just a finite morpism?

Solution 1:

The following property implies what you want.

Let $f : X\to Y$ be a dominant morphism of integral algebraic varieties over $\mathbb C$. Suppose $[K(X): K(Y)]=n$. Then there exists a dense open subset $U$ of $Y$ such that $f^{-1}(y)$ consists in $n$ points for all $y\in U$.

Proof. The first step is to reduce to the case when $f$ is a finite morphism. One can suppose $X=\mathrm{Spec}(B)$, $Y=\mathrm{Spec}(A)$ are affine. The dominant morphism $f$ corresponds to an injective homomorphism $A\to B$. Write $$\mathrm{Frac}(B)=\mathrm{Frac}(A)[t]$$ with $t$ annihilating a polynomial $P(T)\in K(Y)[T]$ of degree $n$ (theorem of primitive element). Replacing $A$ by a localization $A_a$ (geometrically, replace $Y$ by a principal open subset and $X$ by the pre-image of this principal open subset), the element $t$ becomes integral over $A$. As $B$ is a finitely generated algebra over $A$, localizing further $A$, we can suppose $$A\subseteq B\subseteq A[t]$$ (because each element of $B$ belongs to some $A_a[t]$, it is enough to invert a common denominator for a system of generators of $B$ over $A$). As $B$ and $A[t]$ have the same field of fractions and $B$ is finite over $A$, it is easy to see that localizing again $A$, we find $$B=A[t]=A[T]/(P(T)).$$ Now we are almost done. The discriminant $\Delta\in A$ of $P(T)$ belongs to $A$ (we may need to localize $A$ for this) and is non-zero because $P(T)$ is separable in $\mathrm{Frac}(A)[T]$. Let $U$ be the principal open subset $D(\Delta)\subseteq Y$. Then for any $y\in Y$, the fiber $f^{-1}(y)$ is given by the algebra $k(y)[T]/(\bar{P}(T))$ where $k(y)=\mathbb C$ denotes the residue field at $y$ and $\bar{P}(T)\in k(y)[T]$ is the canonical image of $P(T)$. Its discriminant is $\Delta(y)\ne 0$, so it has $n$ (distinct) roots.

Remark. The statement remains true if $Y$ is any integral scheme, $f$ is of finite type, and the extension $K(X)/K(Y)$ is finite separable. But in conclusion, $n$ is the number of points in the geometric fiber $X_{\bar{y}}=(f^{-1}(y))_{\overline{k(y)}}$. The proof is exactly the same. Without the separability hypothesis, the conclusion is for all $y\in U$, $X_y$ is given by a finite $k(y)$-algebra of vector dimension $n$.