TITLE: Degree of a Divisor = Self Intersection?
QUESTION [9 upvotes]: This has been bugging me for some time now. I know that in certain cases that the self-intersection of divisors and it's degree are the same. Like hyerplanes in projective space. I sometimes read that certain degrees are actually defined as the self-intersection.
Is this always the case, are self-intersection and degree basically the same thing? And how are they related?
Thanks in Advance!

REPLY [12 votes]: I think I know what Mike meant to ask. I think the question was: Let $D \subset X$ be a divisor, and embed $X$ into $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ by the standard construction. 

Is the degree of $X$ inside this projective space equal to the
  self-intersection $D^d$?

This is what I have always understood the phrase "the degree of $D$", without further context, to mean. 
As Georges points out, another reasonable interpretation is that we already have some embedding $\phi: X \to \mathbb{P}^N$ and we ask about the degree of $\phi(D)$; that has no relation to the self intersection of $D$.
Now, even with the interpretation above, one needs some caveats. An arbitrary divisor $D$ may not give an embedding to projective space. (Indeed, for an arbitrary $D$, $H^0(X, \mathcal{O}(D))$ may be zero.) But, if $X$ does embed in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ (i.e. if $D$ is very ample), the answer is yes. Correctly formulated, one could probably weaken "very ample" to "ample", or maybe to one of the weaker adjectives like "big" or "nef", but I'll just stick to the very ample case, where there is literally an embedding.
Proof: Let $d= \dim X$. The degree of $X$ in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ is the intersection of $X$ with $d$ generic hyperplanes; $X \cap H_1 \cap H_2 \cap \cdots \cap H_d$. Each those hyperplanes pulls back to a divisor on $X$ which is linearly equivalent to $D$. So $X \cap \bigcap H_i = \bigcap (X \cap H_i)$ and the latter has size $D^d$.