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).


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