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During class today, we were presented quite difficult formulas for computing intersections of divisors.

I'd like to understand this on a more elementary level, what happens when one intersects a degree $0$ divisor $D$ with other divisors/varieties? By intuition I'd say, the resulting divisor has degree $0$ as well?

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    $\begingroup$ What do you mean exactly by "degree zero divisor"? $\endgroup$ – diverietti Jul 26 '12 at 12:22
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    $\begingroup$ A divisor that has $\text{deg }D=0$ $\endgroup$ – Huafeng Kang Jul 26 '12 at 12:23
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    $\begingroup$ Are you kidding? $\endgroup$ – diverietti Jul 26 '12 at 12:28
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    $\begingroup$ But this is a quite special case, since the Picard group of the projective space is $\mathbb Z$. How do you want to define the degree in general? $\endgroup$ – diverietti Jul 26 '12 at 12:45
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    $\begingroup$ Why don't you post one or some of the formulas you want intuition for. Make sure to include all hypotheses on the schemes and other objects in question. $\endgroup$ – name Jul 27 '12 at 1:20
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As has been mentioned, in general 'degree' is not so well/uniquely defined. However, suppose you take a smooth cubic surface in $ \bf{ P}^3$ . There are 27 lines and they should all have degree 1. Take two lines $l_1$ and $l_2$ which are skew and let $D = l_1-l_2$ . This will be 'degree zero' with your definition. Clearly $deg(D|l_1) \neq 0$ Similarly on a quadric hypersurface in $\bf{P}^3$, letting $l_1$ and $l_2$ be two lines which meet , one can construct similar 'examples'. If $X$ is a variety with $Pic(X) = \bf{Z} $ then what you ask is true. I would suspect it fails any other time.

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