# 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.

• 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

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).

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. $\square$

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.
• 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
• 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