# Quail Entry: Degree |D| = Degree of image

I tried to be careful and detailed in my previous post, but it probably wont happen here.

First goal is to explain how for $X \xrightarrow{\phi = |D|} \mathbb{P}^n$

$\deg D = \deg \phi(X)$ (1)

The definition of $\deg \phi(X)$ uses the hilbert function of a projective variety which I wont go into here.  Explanation follows most of ex. II.6.2

The setup: $k = \bar k$, $X = \overline{X} \subset \mathbb{P}^n$ with property that any $Y \subset X$ codim 1 is nonsingular.

### Construct a map. Let $Div(\mathbb{P}^n,X)$ be the free abelian group generated by hypersurfaces $V = Z(F)$ that doesn’t contain $X$.  There is a map

$Div(\mathbb{P}^n,X) \to Div X$

Suffices to define map on $V$, $V \cap X$ breaks up into irreducible components $Y_i$ of codim 1 by

ex I.1.8, ex I.2.6: actually 2.6 is a lengthy problem and the only part of it necessary is that if $Y$ is projective variety and $Y_i = Y \cap U_i \ne \emptyset$, then $\dim Y = \dim Y_i$ where the $U_i = \{x_i \ne 0\}$ are the usual affines of projective space.

so this reduces question to the affine case which ex 1.8 addresses.  Replace $V,X$ with $V_i,X_i \subset \mathbb{A}^n$.  Now $\dim V_i \cap X_i$ = the dimension of a maximal irreducible component, such a component corresponds to a minimal prime $P \subset B = A(X_i) = k[..x_l..]/(g_j)$.

A comm. alg. result says such minimal primes are ht 1 so the dimension is = $\dim Z(P) \cap Z(g_1,.., g_l) = \dim B/P$.  An algebra result says $1 + \dim B/P = ht P + \dim B/P = \dim B = \dim X_i$. Result follows.

So the map is $V \mapsto V.X = \sum v_{Y_i}(f_i) Y_i$ where $f_i$ is a local equation for $V$.

The only choice is local equation for $V$, any two choices differ by a unit which doesn’t matter when you take valuations, so map is well defined by extending by linearity.

Principle goes to Principle. Say $\sum_i n_i V_i = (f)$ and none of the $V_i$ contain $X$; let $f'$ be the restriction of $f$ to $X$. Let $Y_k$ be any codim 1 guy in $X$.  Compare coeff. of $Y_k$ in $(\sum_i n_i V_i).X =: D.X$ with $(f')$.

To accomplish this first describe $(f)$ locally.  Use local equations $f_{ij}$ of $V_i$ in a nbd $U_j$ such that $U_j \cap Y_k \ne \emptyset$.  Thus locally $(f) = \sum_i n_i (f'_{ij}) = (\prod_i f_{ij}^{n_i})$. This is valid in a nbd intersecting $Y_k$ so both functions agree in nbd of the generic point of $Y_k$ therefore

$(f')|_{Y_k} = v_{Y_k}(f') = v_{Y_k}(\prod_i f_{ij}^{n_i})$

$= \sum_i n_iv_{Y_k}(f'_{ij}) = D.X|_{Y_k}$

It follows that get map $Pic(\mathbb{P}^n, X) \to Pic X$

Connect with Intersection Stuff. With notation above, the intersection of $X$ and $V$ along $Y_i$, denoted $i(X,V; Y_i)$ is defined as  length $(S/I_X + I_V)_{P_i}$ over $S_{P_i}$ where $P_i$ is the prime corresponding to $Y_i$.  This sounds less awkward when everything is stated locally.

Goal is to show if $V$ is locally $f_i$ then

$n:= v_{Y_i}(f_i) = i(X,V; Y_i)$.  (2)

Make things local: $\mbox{Spec A}$ is open affine nbd in $X$ of generic point $\eta$ of $Y_i$.  The nonsingularity condition says $O_{X,\eta} \cong A_{P_i}$ is regular of dimension 1. So in it $f_i \mapsto ut^{n}$

Similarly, $(S/I_X + I_V)_{P_i} \cong (A/I_V)_{P_i}$.  Locally can use $I_V = (f_i)$, so localizing get the module in question is $A_p/(ut^{n})$ which clearly has length $n$ over $A_p$: $(t^{n - 1}) \subset (t^{n-1}, t^{n-2}) \subset ... \subset (t^{n-1}, ..., 1)$.

As above write $V.X = \sum_i n_i Y_i$.  Finally, the generalized Bezout Theorem plus (2) gives the latter two equalities in

$\deg V.X = \sum_i n_i \deg Y_i = \sum_i i(V,X; Y_i) \deg Y_i = \deg X \cdot \deg V$

It follows

$\deg D.X = \deg D \cdot \deg X$

Specializing to the case $X$ is a curve and $D$ is a hyperplane, then $D.X$ is a hyperplane divisor, $\deg D = 1$ and so (if $D$ is now the divisor of the map to projective space)

$\deg D = \deg H.X = 1\cdot \deg X$

which gives (1).