# Fibre cardinality of an unramified morphism

Let $\varphi: X \to Y$ be a finite, dominant, unramified morphism of varieties over an algebraically closed field. If necessary, we can assume $X$ and $Y$ to be nonsingular. I am trying to prove that

$$\mathrm{deg}(\varphi):=[K(Y):K(X)]=|\varphi^{-1}(P)|$$

for every point $P\in Y$. The statement is very easy to prove for curves. However, I am completely stuck trying to prove it for higher dimensions. I cannot find this statement or a similar one in literature, it would also be a great help if someone could point me there.

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The field extension doesn't make sense unless $\varphi$ is dominant. Even so, it clearly fails if $\phi$ is an open immersion and $X\neq Y$. So, what do you asssume exactly? –  Laurent Moret-Bailly Jan 20 '12 at 16:27
You may compute $X_y={\rm Spec}\ (\phi_*({\cal O}_X))_y$ and $(\phi_*({\cal O}_X))_y$ is a vector space of dimension $r$ over $\kappa(y)$; it is also an (étale !) algebra over $\kappa(y)$ and hence it must be a direct sum of separable extensions of $\kappa(y)$. So if $\kappa(y)$ is alg. closed (so that it has only the trivial separable extension) then it must be a direct sum of $r$ copies of $\kappa(y)$, the direct sum being viewed as a $\kappa(y)$-algebra. –  Damian Rössler Jan 20 '12 at 16:55
A finite dominant map between non-singular varieties is automatically flat. This follows from the fact that a finite injective map between regular local rings is flat. –  Keerthi Madapusi Pera Jan 20 '12 at 17:38
Note that my comments above are strictly based on the hypotheses (no regularity hypothesis is made on $X$ or $Y$). –  Damian Rössler Jan 20 '12 at 18:04
@Keerthi: Re: "In fact, more is true. If you have a map $f:X\to Y$ of irreducible varieties with the target smooth, and with all fibers of $f$ equi-dimensional of dimension $\dim X−\dim Y$, then $f$ is flat. See EGA IV.6.1.5.". This is actually not true as stated. You need $X$ to be Cohen-Macaulay. Check EGA IV.6.1.5. (It's not just that you need this because it is stated, but the statement is not true otherwise. If $f:X\to Y$ is finite and $Y$ is non-singular, then $f$ is flat iff $X$ is Cohen-Macaulay. –  Sándor Kovács Jan 21 '12 at 0:22

## 2 Answers

After I wrote the comments above, I found the following reference :

Formula (12.6.2), p. 329 in Görtz-Wedhorn, Algebraic Geometry I, Viehweg & Teubner Verlag

for (a generalisation of) the equality you are looking for, when $\phi$ is assumed flat (which is true if you assume that $X$ and $Y$ are non-singular, as pointed out in the comments of K. M. Pera and S. Kovacs).

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Claim Let $\phi:X\to Y$ be a finite étale morphism (i.e., flat and unramified) of reduced schemes of finite type over an algebraically closed field. Assume that $Y$ is irreducible and Cohen-Macaulay and $\phi$ is dominant on every irreducible component of $X$. Then for a closed point $P\in Y$, the number of pre-images of $P$, denoted by $|\phi^{-1}(P)|$ is independent of $P$. Define $\deg\phi$ to be this value. If $X$ and $Y$ are both irreducible, this value is equal to $[K(Y):K(X)]$.

Proof Since $Y$ is connected, the statement is local and we may assume that $Y$ is affine and hence quasi-projective. Let $P\in Y$ and if $\dim Y>1$, then let $H\subseteq Y$ be a general effective very ample divisor such that $P\in H$. By the assumptions $H$ is again an irreducible reduced Cohen-Macaulay scheme of finite type over an algebraically closed field. Replace $Y$ with $H$ and $X$ with $\phi^{-1}H$. Notice that the original assumptions remain true, in particular $\phi^{-1}H\to H$ is étale and for any $P,Q\in Y$ and any two general effective very ample divisors $H_P,H_Q\in Y$ such that $P\in H_P$ and $Q\in H_Q$ it follows that $H_P\cap H_Q\neq\emptyset$.

Therefore we may assume that $\dim Y=1$. It follows that $\dim X=1$ and since $\phi$ is étale, the irreducible components of $X$ are disjoint, so we may assume that $X$ is also irreducible. Let $\widetilde Y\to Y$ be a resolution of singularities of $Y$ and consider the base change $\widetilde\phi: \widetilde X\to \widetilde Y$. Since $\phi$ is étale, so is $\widetilde\phi$ and hence $\widetilde X$ is also non-singular. In other words we may assume that $X$ and $Y$ are nonsingular curves. In that case $\{P\}\subset Y$ is a divisor and it is well-known that $\deg\phi^*P=[K(Y):K(X)]$. We obtain that $\deg\phi^*(P)$ is independent of $P\in Y$ for arbitrary points and since $\phi$ is étale, this implies that if $P\in Y$ is a closed point, then $|\phi^{-1}(P)|=\deg\phi^*(P)$ is independent of $P$.

Now if $X$ and $Y$ are both irreducible at the start, then since the value $|\phi^{-1}(P)|$ is independent of $P$ for closed points, it is enough to check that value at one particular point. As $\phi$ is étale, the field extension $K(Y)\subseteq K(X)$ is (finite) separable and hence may be generated by a single element with a minimal polynomial of degree $[K(Y):K(X)]$. This shows that $Y$ may be embedded in some projective space $\mathbb P^N$ such that $X$ is birational to a hypersurface in $\mathbb P^N\times \mathbb A^1$ such that (as a rational map) $\phi$ is the composition of the projection to $\mathbb P^N$ and the birational map on $X$. It follows that for a general (closed) point the equality $|\phi^{-1}(P)|=[K(Y):K(X)]$ holds. $\square$

Comments
1) Perhaps someone has a simpler argument for the last paragraph, but I don't see how one can easily relate the degree of the field extension to the number of points in a fiber. The appearance of this field extension degree in the proof of the first statement would need further work, because it is the degree of the field extension of the complete intersection curves we obtain by taking hyperplane cuts. Maybe it is obvious, but I am not sure how to prove easily that the degree of that field extension is the same as the degree of the original. One way to prove it is a variation of the last paragraph above.
2) Obviously the statement is only for closed points, but I am sure that's what Jesko meant.
3) Just for the record: the above statement implies the one in the question in case $X$ is Cohen-Macaulay and $Y$ is non-singular as those together imply that $\phi$ is flat.
4) See alternative proof in the comments above by Damian Rössler.

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The quantity $\#\phi^{-1}(P)$ is independent of $P$ (for $P$ closed) because it is the rank of $\phi_*({\cal O}_X)$ (see my comments above). Note that $\phi_*({\cal O}_X)$ is flat and coherent and hence locally free, because $Y$ is a noetherian scheme and $\phi$ is flat and finite (and in particular affine). I dont understand why you think that it is necessary to reduce to the case of curves (?) –  Damian Rössler Jan 21 '12 at 10:04
Dear Damian, first of all, thank you for your comment. It is always good to know that someone else actually reads what one writes. Second, I do not understand why you think that I think that it is necessary to reduce to the case of curves. I do think that such a reduction makes for a very simple proof. I would argue that the advantage of this proof is that it uses very little. The reduction step is nearly trivial, and so is the curve case. (cont'd) –  Sándor Kovács Jan 21 '12 at 21:23
Yes, $\phi_*\mathscr O_X$ is indeed locally free, but to make the further conclusions you would likely have to appeal to more advanced results. (I may be wrong, but I think this is actually irrelevant) In particular, I did not see a proof of the fact that the desired value is equal to the field extension. (I don't doubt that you can prove it. I am just saying I did not see it even mentioned.) (cont'd) –  Sándor Kovács Jan 21 '12 at 21:23
An important point here is that the original question regards one of those statements that "is obvious" and one might use without proof in an argument. When one wants to prove such a statement it is preferable if one does not use machinery that is more complicated than the question at hand. Of course, you may argue (and perhaps this is what you are saying) that your proof is simpler than mine. Fine. However, I fail to see how that opinion of yours should imply that I should not post my own proof if I like. (cont'd) –  Sándor Kovács Jan 21 '12 at 21:24
Finally, your comment has not much to do with my answer. It seems to suggest that you consider this a kind of competition. I don't. –  Sándor Kovács Jan 21 '12 at 21:24