There is a more geometric way to explain Felipe's answer. To compute the degree of a closed subvariety $Y$ of projective space, of dimension $n$ say, you intersect it with $n$ different hyperplanes in sufficiently general position (with respect to each other and also with respect to $Y$), so that the result is a collection of points, and you add up the number of points. This is the degree of $Y$.
In other words, degree has three properties that serve to define it:
(a) it is additive with respect to unions (EDIT: of distinct varieties of the same dimension, say, to avoid scheme-theoretic issues).
(b) the degree of a point is one.
(c) degree is preserved by taking sufficiently general hyperplane sections.
(I could omit the caveat sufficiently general here is I was willing to work
with scheme structures, and not just varieties).
(It is then an exercise, using these conditions, to relate the degree as defined this way
to the degree defined via the Hilbert polynomial.)
Now if $Y$ is the image of $X$ under the emdedding given by $|m A|$, then the intersection
of a hyperplane with $Y$ pulls back, under the isomorphism $X buildrel sim over to Y$,
to a member of the linear system $|m A|.$ (This is by the very definition of the embedding
given by $|m A|$.) When you intersect $n$ such divisors, the number of points you get is
then (m A)^n. (Because intersection is invariant under deformation of the cycles being
intersected, you can replace all the divisors by the linearly equivalent divisor $m A$.)
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