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I've seen divisors on curves before, a few years ago in a course in algebraic geometry. Now I've come across them again, but they're somewhat more generalized. I was hoping someone could explain the similarities/differences between the two notions.

The first notion (from Algebraic Curves by Fulton) says for any irreducible projective curve $C$, with nonsingular model $X$, a divisor on $X$ is a formal sum $$ D = \sum_{P\in X} n_P P$$ where each $n_P\in \mathbb{Z}$ and all but finitely many are 0.

The second notion (from Algebraic Geometry by Milne) says for any normal, irreducible variety $V$, a divisor on $V$ can be written uniquely as a finite (formal) sum $$ D = \sum n_i Z_i,$$ where the $Z_i$ are the irreducible subvarieties of $V$ of codimension 1 (prime divisors).

In the generalized case, when $V$ is a curve (i.e., dimension 1) then the $Z_i$ are 0 dimensional, so just points on the curve, and everything seems OK. But then we move on to the degree of a divisor. In Fulton, the degree of a divisor $\sum_{P\in X} n_P P$ is simply the sum of the coefficients, $\sum_{P\in X} n_P$. However, via Example 12.4 in Milne, the definition of a divisor is as follows:

Let $C$ be a curve. If $D = \sum n_i P_i$, then the intersection number $$ (D) = \sum n_i[k(P_i):k].$$ By definition, this is the degree of $D$.

Now, we don't have any references for my class (as it's all over the place and would require about 10 minimum..), but this is how we defined the degree of a divisor for a curve in class (using the extension of $k(P_i)$ over $k$) and I'm confused. Is this suggesting that, by Fulton's definition, we have $[k(P_i):k] = 1$ for every point? Because if so, it seems silly to even bother putting that in. Now given Milne is actually writing the intersection number, I felt he could just leaving implicit that the $[k(P_i):k]$ are 1. However in class, we simply said a divisor is blah and its degree is blahh using the degree of the extensions, so there must be something non-trivial going on here!

Now the only thing I haven't entirely accounted for above, is the fact that in Fulton, a curve over a field $k$ is a set of points in $k^n$, whereas Milne is using algebraic varieties, which I'm not terribly well versed with. However, Milne makes note of the fact that there is a one-to-one correspondence between maximal ideals of $k[V]$ and one point sets of $V$, so I would expect, for each $P_i$ in Fulton's definition, there is a corresponding maximal ideal, so that a divisor should be identical, except replacing each $P_i$ with the corresponding maximal ideal.

Somewhat more troubling, I thought it was obvious at first, and didn't give it a second thought, but in retrospect, what is $[k(P_i):k]$? If $P_i$ is a subvariety of codimension 1, for a curve defined over $k$, then how can this not have to be 1? If nothing else, I'd also appreciate any additional references for these generalized divisors on curves (though I do like Milne's notes!). Thanks!

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This is related to the fact that not every closed point is rational. For example, Fulton might assume you are over $\mathbb{C}$ or at least an algebraically closed field and hence $k(P_i)=k$, so the degree is always 1. In general, you only know that the local ring at a point modulo the maximal ideal is an algebraic extension of the base field, which may be non-trivial. If no one has posted an expanded version of this comment when I wake up tomorrow I'll expand it. –  Matt Feb 25 '12 at 7:27
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In a nutshell, on page 35 Fulton states in the first sentence of the chapter "From now on $k$ will be a fixed algebraically closed field". This implies that all $[k(P_i):k]=1$ so that his definition is compatible with yours and Milne's.

In more detail, Milne also supposes that his base field is algebraically closed in the first 10 chapters but in Chapter 11 he introduces varieties over an arbitrary field $k$.
This means that an affine varietie $V$ is determined by some finitely generated $k$-algebra $A$ and consists of the maximal ideals of $A$.
In practice $A=k[T_1,...,T_N]/\langle P_1, ... ,P_r\rangle$ and we say that $V$ is the subvariety $V\subset \mathbb A^N_k$ determined by the equations $P_j\in k[T_1,...,T_N]$.
Given a point $P\in V$ corresponding to a maximal ideal $\mathfrak m\subset A$, the field $\kappa(P)=A/\mathfrak m$ is a finite-dimensional extension of $k$ (this is a version of the Nullstellensatz and is not trivial) and its dimension $[\kappa(P):k]$ is the integer that you asked about
(I prefer to write $\kappa(P)$, as in EGA and many texts in algebraic geometry, instead of $k(P)$ so as to avoid confusion with the base field $k$)

An example Suppose you take $A=\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle$, that is you consider the circle $V\subset \mathbb A^2_{\mathbb R}$ of equation $X^2+Y^2-1=0$.
If you consider the maximal ideal $\mathfrak m=\langle x-1,y \rangle\subset A=\mathbb R[x,y]$ corresponding to the good old point $P=(1,0)$ of the circle, you get the unsurprising equality $[\kappa(P):\mathbb R]=1$, since $\kappa(P)=\mathbb R[x,y]/\langle x-1,y\rangle=\mathbb R$.
However if you consider the point $Q$ corresponding the maximal ideal $\mathfrak n=\langle x-\sqrt 2,y^2+1\rangle\subset A=\mathbb R[x,y]$, you get $[\kappa(Q):\mathbb R]=2$ since $\kappa(Q)=\mathbb R[x,y]/\langle x-\sqrt 2,y^2+1\rangle\cong \mathbb C$.

Edit
Since Alex asks about this in his comment below: yes there is a relationship between the point $Q\in V$ and the two points $( \sqrt 2,\pm i)$ in the complexified circle $V_\mathbb C\subset \mathbb A_\mathbb C^2$ corresponding to the $\mathbb C$-algebra $A\otimes_\mathbb R \mathbb C=\mathbb C[X,Y]/\langle X^2+Y^2-1\rangle$.
You can say that $\mathfrak n$ corresponds to the orbit of either of these points under the action of the Galois group $Gal(\mathbb C/\mathbb R)$ on $V_\mathbb C$ or you can also say that $\mathfrak n$ is the common kernel of the two homomorphisms of $\mathbb R$-algebras $A\to \mathbb C$ sending $x\mapsto \sqrt 2$ and $y\mapsto \pm i$

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!! That explains it! I looked through Chapter 8 of Fulton, along with all the areas he refers to for previous results, and couldn't find anything about $k$ being algebraically closed, but that settles that. Small question about the last part of the example, what is the point $Q$? If instead, $\mathfrak{n} = \langle x, y^2 + 1\rangle$ I'd say this corresponds to the point $(0,i)$, but the point $(\sqrt{3},i)$ isn't on the circle $V$, so am I missing something silly? Thanks! –  Alex Feb 25 '12 at 21:11
    
Dear @Alex, there was indeed a typo in my post : I meant $\sqrt 2$ where I had written $\sqrt 3$. I just corrected that: thanks a lot for being so attentive. I have also added a few words on the link with the complexified circle in an edit. This is a vast subject belonging to arithmetic geometry. The key concept you might look up some time would be "closed points versus rational points". –  Georges Elencwajg Feb 25 '12 at 22:07
    
Ah, great! Thanks again, and thanks for the extra bit in the edit (and reference :) )! –  Alex Feb 25 '12 at 22:31
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