I've become extremely confused (due to having no experience with varieties) over a remark (Remark 3.7) Silverman makes in his book The Arithmetic of Elliptic Curves. Here's the relevant background:
If $\phi: C_1 \rightarrow C_2$ is a nonconstant morphism of smooth curves over $K$ then the pullback map $\phi^* : K(C_2)\rightarrow K(C_1)$ extends to a homomorphism on divisors $\phi^*:\text{Div}(C_2)\rightarrow\text{Div}(C_1)$ given by $\phi^* (Q) = \sum_{P\in \phi^{-1}(Q)} e_\phi (P) (P)$ and extended $\mathbb Z$-linearly to all divisors. There is also the notion of a pushforward homomorphism of divisors $\phi_*: \text{Div}(C_1)\rightarrow\text{Div}(C_2)$ given by $\phi_* (P) = (\phi(P))$, similarly extended by linearity. These morphisms satisfy some nice properties; I think the relevant ones are
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$\deg (\phi^* D) = \deg \phi \deg D$ for all divisors $D\in \text{Div}(C_2)$
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$\phi^* \text{div}(f) = \text{div}(\phi^* f)$ for all $f\in K(C_2)$
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$\deg \phi_* D = \deg D$ for all $D\in \text{Div}(C_1)$
Silverman then claims that because these properties imply that $\phi_*, \phi^*$ send divisors of degree zero to divisors of degree zero they extend to homomorphisms on the Picard groups – $\phi_* :\text{Pic}(C_1)\rightarrow \text{Pic}(C_2)$ and $\phi^* : \text{Pic}(C_2)\rightarrow \text{Pic}(C_1)$.
Now he considers the case where we have a morphism $f: C\rightarrow \mathbb P^1$, which we can also think of as an element of $K(C)$. He claims that
$\deg \text{div}(f) = \deg f^* ((0)-(\infty)) = \deg f – \deg f = 0$
I understand why $\deg f^* ((0)-(\infty)) = 0$, as I believe it follows from these Picard homomorphisms, but I'm struggling to "see" the first equality; why should $\deg \text{div}(f) = \deg f^* ((0)-(\infty))$?
Working everything through, if I haven't made a mistake we get
$\deg \text{div} (f) = \sum_{f(P)=0} \text{ord}_P (f) – \sum_{f(P) = \infty}\text{ord}_P (f)$;
$\deg f^* ((0)-(\infty)) = \sum_{f(P)=0} e_f (P) – \sum_{f(P) = \infty}e_f(P)$
It then seems reasonable that to prove these two things are equal we should prove that for a zero (respectively a pole) $P\in C$ of $f$ we should have $\text{ord}_P (f) = e_f (P)$. Let $t_0\in K(\mathbb P^1)$ be a uniformiser for $f(P) = 0$; then $e_f (P) = \text{ord}_P (f^* t) = \text{ord}_P (t\circ f)$. So I want to be able to prove that $\text{ord}_P (f) = \text{ord}_P(t\circ f)$ for any uniformiser $t$ of $f(P)$ (the little intuition I have in this area tells me this ought to be true). But I'm really struggling to prove this formally using the maximal ideals of the local rings $K[C]_P$ and $K[\mathbb P^1]_{f(P)}$. Is this true, and can you provide a proof?
I suppose the question can be phrased more generally as follows: given morphisms of smooth curves $C\xrightarrow{f}\mathbb P^1 \xrightarrow{g} \mathbb P^1$, it makes sense to consider $f\in K(C), g\in K(\mathbb P^1)$. Then it would be nice if the valuation behaved multiplicatively on composition i.e. $\text{ord}_P (g\circ f) = \text{ord}_P (f) \text{ord}_{f(P)} (g)$. Again I believe this intuitively but have no real idea how to phrase the proof.
Edit: I am also aware that $\text{ord}_P (\phi^* g) = e_g (P)\text{ord}_{\phi(P)} (g)$ in the general case above (where $g\in K(C_2)$ and $P\in C_1$). However, in the case $C_2 = \mathbb P^1$, $\phi = f$ and $g = t$ is a uniformiser, this gives no new information.
Best Answer
Let $(A,\mathfrak{m})$ denote the local ring of $C$ at $P$, with uniformizing parameter $s\in\mathfrak{m}$. Let $f$ be a function on $C$, which is regular at $P$. Let $Spec(A')$ be an open affine neighborhood of $P$ which does not meet the locus which maps to $\infty$. Then, we can view $f$ in two ways:
(1) $f$ is an element of $A'$ and hence defines an element of $A$. By definition of $ord$, up to a unit in $A$, this element is $s^{ord_P(f)}$.
(2) $f$ is a function, $f:Spec(A')\rightarrow \mathbb{A}^1$. Equivalently, $f$ is a map $f^*:k[t]\rightarrow A'$. By definition, it is the map which sends $t$ to the regular function $f$ on $SpecA'$, i.e. $f^*t=f$ where the right hand side is in the sense of (1).
So it's confusing because $f^*t=f$ seems to make no sense. But when we say $f^*$ on the LHS, we're viewing $f$ as a map of varieties, and when we say $f$ on the RHS, we're viewing $f$ an element of the ring of regular functions at $P$.
Viewed another way, it's the only thing that could possibly make sense. Whenever you have a map to $f:X\rightarrow\mathbb{A}^1$, you get a regular function on $X$ by pulling back the regular function $t$ on $\mathbb{A}^1$. What else could that regular function be but $f$?