# 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?

• 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? – Ehsan M. Kermani Mar 26 '13 at 2:04
• These are related recent discussions, in case you haven't seen them math.stackexchange.com/questions/340565/… and math.stackexchange.com/questions/340687/… – Ehsan M. Kermani Mar 26 '13 at 2:07
• Your phrase "More generally..." is not warranted since an injective morphism needn't be finite: think of the injection $\mathbb A^1\hookrightarrow \mathbb P^1$ – Georges Elencwajg Mar 26 '13 at 7:11
• 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. – Jesko Hüttenhain Mar 26 '13 at 7:22
• @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. – Li Yutong Mar 26 '13 at 16:21

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 $$n$$.

• 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. – Li Yutong Apr 8 '13 at 1:19
• 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! – Georges Elencwajg Sep 1 '13 at 9:19
• @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. – Georges Elencwajg Sep 1 '13 at 20:47
• @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? – Manos Apr 13 '17 at 14:47
• 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 :-( – Georges Elencwajg Apr 13 '17 at 18:33