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Let $f: X \rightarrow Y$ be a finite morphism of schemes, defined here,

I always assumed that the degree of $f$ was the degree of the induced field extension $$ [K(Y):K(X)], $$ of course only defined if $f$ is dominant (hence surjective).

However, is this standard?

Or is the degree of $f$ defined as: Let $f$ locally be given by maps of rings $B_i \rightarrow A_i$. The degree of $f$ is the minimal number of elements of $A_i$ that generate it as a $B_i$ module?

I'm almost sure this does not work.

Q1: Why not? (excuse my lack of commutative algebra knowledge) (i am more than happy with just a sketchy answer to this!)

Q2: Does it work if we assume $X$ and $Y$ to be varieties over $\mathbb{C}$? If so/not, why?

Q3: What is indeed the commonly agreed definition of the degree of a finite map?

That's quite a lot, thanks for your answer!

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When you say "locally", do you mean "there exists an affine covering such that ...." or "for any affine covering ...." ? Do you take the minimum of the "local degrees" ? – user18119 Mar 25 '13 at 12:05
up vote 3 down vote accepted

I just want to note that you don't need any finiteness condition on $f$ to define the degree. Any dominant morphism $f : X \to Y$ of integral schemes $X$ and $Y$ maps the generic point $x$ of $X$ to the generic point $y$ of $Y$, hence there is a canonical homomorphism $\mathscr{O}_{Y,y} \to \mathscr{O}_{X,x}$, which gives a field extension $R(Y) \hookrightarrow R(X)$. (Thanks for Georges Elencwajg for simplifying my original argument.)

So we don't need finiteness conditions on $f$, or on $X$ and $Y$, to talk about the field extension $R(Y) \hookrightarrow R(X)$ or its degree $\deg(f) = [R(X) : R(Y)]$. However if one wants to ensure this degree is finite, it suffices to assume $f$ is locally of finite type and

(i) $f$ is closed, the dimensions of $X$ and $Y$ are equal (see (Stacks, 02JX) and (Stacks, 02NX)),


(ii) $f$ is separated or quasi-compact (e.g. proper or even just universally closed), and there exists an affine open $V \subset Y$ such that the restriction $f^{-1}(V) \to V$ is finite (Stacks, 02NX).

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Adeel, so if i understand correctly the notion of degree of a map and finiteness of a map are two independent ones? Which just happen to be very useful together as in Georges' answer? – Joachim Mar 25 '13 at 14:20
Yes, finiteness is just an easy way to ensure that the degree is finite, but it is not necessary. – Adeel Mar 25 '13 at 14:21
Thanks Adeel, that helps. – Joachim Mar 25 '13 at 14:22
About Q1 and Q2, so the degree is never ever defined using the rank of the module as in my question? – Joachim Mar 25 '13 at 14:23
It is possible that it coincides with that definition in some special case, but I am not sure. – Adeel Mar 25 '13 at 14:27

If you want to talk of function fields, it is prudent to consider only finite dominant morphisms $f\colon X\to Y$ between (locally noetherian) integral schemes.
The degree of $f$ is then defined as the degree of the corresponding extension of function fields $n=[k(X):k(Y)]$.
Qing Liu in his wonderful book asks you to prove (Chapter5, Exercise 1.25) that if $f$ is flat, then all fibers $f^{-1}(y)$ of $f$ have $n$ points, if you count them suitably: $\dim_{k(y)}\mathcal O(f^{-1}(y))=n$ .

Warning: The result is always false if $f$ is not flat!
This is part of the exercise mentioned above.
For example if you normalize the node $N=V(y^2-x^2-x^3)\subset \mathbb A^2_k$, you obtain the finite morphism $$f\colon \mathbb A^1_k\to N\colon t\mapsto (t^2-1,t^3-t)$$ of degree $$[k(\mathbb A^1_k):k(N)]=[k(t):k(t^2-1,t^3-t)]=[k(t):k(t)]=1$$
However for the fiber over $P=(0,0)\in N$ you have $\dim_{k(P)}\mathcal O(f^{-1}(P))=2$ because $f^{-1}(P)$ consists of the two simple points $t=-1, t=+1$.

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Georges, i understand that the degree is not defined for a finite morphism of varieties over $\mathbb{C}$ that is not dominant? (excuse the naive general use of schemes by the way, as you can probably tell i am too used to working with schemes with nice properties) – Joachim Mar 25 '13 at 13:36
Yes, if a morphism is finite but not dominant, it is better not to assign a degree to it . For example the morphism $f:\mathbb A^1\to \mathbb A^2: t\mapsto (t,0)$ is finite, but is not dominant and does not induce a morphism of fields $k(T_1,T_2) \to k(T_1)$. So $f$ has no degree. – Georges Elencwajg Mar 25 '13 at 13:59
Georges, thanks a lot for the context and reference to the book. Since the other answer was more specifically about the notion of degree, i decided to accept that one, but your answer was definitely useful too! – Joachim Mar 25 '13 at 14:30
That's quite all right Joachim: you are the only one capable of telling which answer best solves your problem. And, anyway, it gave me great pleasure to answer your interesting question! – Georges Elencwajg Mar 25 '13 at 14:39

Besides Adeel's nice answer, one can find another definition of the degree in case the morphism is flat and finite (and the schemes are locally Noetherian) in the Stacks project 24.46.1 and 24.46.2. This is actually the thing i hoped it would be, as i stated in the question.

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