I've always thought of the degree of a subvariety of projective space as the degree of the divisor that defines the (given) embedding into projective space. It's been pointed out to me that this works only for curves. Now I'm confused: is there a similar characterization of the degree of a general subvariety of some projective space?
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1$\begingroup$ Maybe the reason for your confusion is that on a curve the degree of a divisor is intrinsically defined, whereas in general the notion of degree of a divisor depends on the projective embedding of the variety. Nevertheless, as Allen says below, the relationship still holds in higher dimensions. $\endgroup$– JohanCommented Mar 27, 2010 at 18:28
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3 Answers
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If $X\subset \mathbb P^n$ is a subvariety of dimension $m$ embedded by a linear system $V \subset H^0(X,\mathcal O_X(D))$ then the degree of $X$ is equal to $D^m$.
answered Mar 26, 2010 at 4:17
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2$\begingroup$ To elaborate on jvp's point slightly, you can't even define the degree of a divisor on a variety that isn't a curve. On a curve, you count the points. But what about on a surface? You would need the degrees of curves, which are less obvious. However, the self-intersection of a curve is well defined and gives the right answer. $\endgroup$ Commented Mar 26, 2010 at 11:13
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See the section about intersection of chapter I of hartshorne. Intuitively degree is the number of intersection of the dimension r subvariety and the GENERIC dimension n-r linear subspace if they are both in P^n. A rigorous definition was given in that section by the leading coefficient of the Hilbert polynomial.
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I disagree with "this works only for curves". Say we've already defined degree for schemes up to dimension $n$. Then use your rule to define it for schemes of dimension $n+1$. Dr. "This works only for curves" is unwilling to go beyond the $n=0$ case, for some reason.
answered Mar 26, 2010 at 16:36