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

http://en.wikipedia.org/wiki/Finite_morphism

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!

Joachim
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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" ? –  Mar 25 '13 at 12:05

3 Answers3

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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 $t=-1, t=+1$.

Georges Elencwajg
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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
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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)),

or

(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).

  • 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. –  Mar 25 '13 at 14:21
  • 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. –  Mar 25 '13 at 14:27
  • I assume that that means it is indeed far from commonly done and i should not bother. Thanks again! – Joachim Mar 25 '13 at 14:29
  • It's more probably a consequence of me working in far too much generality without understanding more concrete things first :) –  Mar 25 '13 at 14:33
  • Dear Adeel, your corollary is false: all homomorphisms of fields are injective but many morphisms of integral schemes are not dominant. – Georges Elencwajg Mar 25 '13 at 14:44
  • Ah nice, i already wondered about that! – Joachim Mar 25 '13 at 14:45
  • Ah sorry, I miscopied this from my notes. Good catch! And you mean nontrivial homomorphisms of course. –  Mar 25 '13 at 14:47
  • Dear Adeel, I am sorry to come back to that corollary, but your modified version makes no sense since in general there is no morphism $R(Y)\to R(X)$, as shown by the example $\mathbb A^1\to \mathbb A^2$ in my answer. So there is no general notion of degree for non finite morphism. – Georges Elencwajg Mar 25 '13 at 14:57
  • @GeorgesElencwajg I don't understand. As you said, your morphism $\mathbb{A}^1 \to \mathbb{A}^2$ is not dominant. Dominant morphisms map generic points to generic points and there is always a canonical homomorphism $\mathscr{O}{Y,y} \to \mathscr{O}{X,x}$. –  Mar 25 '13 at 15:00
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    Dear Adeel, sorry, you are right. The corollary is true but doesn't say very much since all morphisms of fields are injective anyway. (I really mean all morphisms of fields: the zero map is not a field morphism because it doesn't send $1$ to $1$). And also, let me tell you that I really hate criticizing another answer when I wrote one myself: be assured that I only did so because there was a factual little problem. – Georges Elencwajg Mar 25 '13 at 15:17
  • Good point. We don't need this corollary then, we just use the dominance of $f$ to get the field extension, and anyway we still get a well-defined degree for dominant morphisms (just in an easier way). Thank you very much for your comments. –  Mar 25 '13 at 15:34
  • @GeorgesElencwajg and Adeel. Thanks for the discussion, reading such conversations is precisely why i love this site. Another question came up though. In the last comment to the accepted answer to http://mathoverflow.net/questions/72364/finite-tale-morphism-of-varieties/72366#72366 Anton Geraschenko does use the fact that if the degree of the map is $n$, the map is locally given by an inclusion of rings $R \rightarrow S$ which makes $S$ into a free (!) module of rank $n$. The map here is assumed to be finite surjective étale. So this is "some special case", as you called them earlier Adeel? – Joachim Mar 26 '13 at 17:11
  • Oh wait, also the varieties are assumed to be nonsingular, and over an algebraically closed field.. – Joachim Mar 26 '13 at 17:38
  • @Joachim, I'm not sure, I don't have much experience with finite morphisms. Perhaps Georges knows. –  Mar 26 '13 at 17:52
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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 02K9. This is actually the thing I hoped it would be, as I stated in the question.

Armando j18eos
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Joachim
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