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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?

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  • $\begingroup$ Identify $X$ with its image in $Y.$ Do you see now why they have equal function fields? do they have common (isomorphic) open subsets for example? $\endgroup$ Mar 26, 2013 at 2:04
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    $\begingroup$ These are related recent discussions, in case you haven't seen them math.stackexchange.com/questions/340565/… and math.stackexchange.com/questions/340687/… $\endgroup$ Mar 26, 2013 at 2:07
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    $\begingroup$ Your phrase "More generally..." is not warranted since an injective morphism needn't be finite: think of the injection $\mathbb A^1\hookrightarrow \mathbb P^1$ $\endgroup$ Mar 26, 2013 at 7:11
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    $\begingroup$ You only get a morphism $K(Y)\to K(X)$ if $\pi$ is also dominant. If it is dominant and injective, it identifies $X$ with some dense subset of $Y$, and they therefore share an open affine. How do you define finite morphism? That'd be good to know in order to answer your second question. $\endgroup$ Mar 26, 2013 at 7:22
  • $\begingroup$ @Ehsan M.Kermani, wow, great! I did not notice a similar question just posted yesterday! That helps me clarify how to define the degree of the morhpism, but I still want to keep my question here because it asked the correspondence between degree and points in the fibre. $\endgroup$
    – Li Yutong
    Mar 26, 2013 at 16:21

1 Answer 1

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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 $n$.

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  • $\begingroup$ Thank you for your beautiful explanation!!! I was wondering if there is any easy way to justify the claim "The first step is to reduce to the case when $f$ is a finite morphism." To me, this seems the Zariski's main theorem in the sense of Grothendieck. $\endgroup$
    – Li Yutong
    Apr 8, 2013 at 1:19
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    $\begingroup$ Dear user 18119: I'm sad that you no longe post under your real name. But I was still sadder to read in your user page the words "please delete me". Please, please, I implore you dear user: stay with us! $\endgroup$ Sep 1, 2013 at 9:19
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    $\begingroup$ @user 18119: sad news , but thanks for all the brilliant contributions from a great expert . I wish you all the best, professionally and personnally: godspeed, dear friend. $\endgroup$ Sep 1, 2013 at 20:47
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    $\begingroup$ @GeorgesElencwajg: I think i figured it out:Since $A[t]$ is integral over $A$, so is $B$. Because $B$ is a finitely generated algebra over $A$, which is also integral over $A$, it is a finitely generated $A$-module, say $B = A b_1+\cdots + A b_n$. It follows that $K(B) = K(A)[b_1,\dots,b_n]$. Since $K(B) = K(A)[t]$, we have that $t$ is a polynomial in the variables $b_i$ with coefficients in $K(A)$. Hence localizing $A$ at finitely many elements we have that $t$ is inside $A[b_1,\dots,b_n]$, the latter of course being equal to $B$. Thus $A[t] \subset B$ and so $B = A[t]$. Do you agree? $\endgroup$
    – Manos
    Apr 13, 2017 at 14:47
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    $\begingroup$ Dear @Manos: I read your comment with pleasure and it looks quite correct. However I would have to really immerse myself again in the situation in order to give a 100% guarantee. Sorry about that :-( $\endgroup$ Apr 13, 2017 at 18:33

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