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This has been bugging me for a while now.

Say I have a projective variety given by some polynomial $P$ and the canonical divisor of the projective space. How can I concretly calculate the Intersection of the two?

And by concretly I mean, actually get a result that is not purely abstract? (Like actual Intersection points, degree, etc...)

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If I'm interpreting your question correctly, then you have two homogeneous ideals and can add them together to get the intersection of your two subschemes. Perhaps I am completely missing the point of your question though.. – jrajchgot Jul 1 '12 at 23:42
up vote 5 down vote accepted

Unfortunately, you will run into a basic difficulty in your desire to make this example concrete, which is that the canonical bundle on projective space is anti-ample, and so the canonical bundle is not effective. (Rather, it is its dual that is an effective divisor.)

More precisely, if $H$ is the linear equivalence class of a hyperplane in $\mathbb P^n$, then the canonical divisor $K$ is equal to $-(n+1)H$.

So if you want to intersect $K$ with $V(P)$ (the variety cut out by the polynomial $P$) you will have to use at least a little bit of theory, even if only to interpret what you are doing.

You might be better of starting with $H$ itself. Then $H \cap V(P)$ is the linear equivalence class of a hyperplane section of $V(P)$. Assuming that $V(P)$ is smooth, then Bertini's theorem says that a generic member of the linear equivalence class $H\cap V(P)$ will be smooth, and even irreducible if the dimension of $V(P)$ is at least two (i.e. if $n \geq 3$).

Then one way to write $K \cap V(P)$ is simply as $-(n+1) \bigl( H \cap V(P) \bigr)$.

Alternatively, one could consider the linear equivalence class $(n+1)\bigl( H \cap V(P)\bigr).$ This is the class of intersections of $V(P)$ with a degree $n+1$ hypersurface, and again a generic member is smooth (and irreducible if $V(P)$ is of dimension at least $2$). Then you think of $K \cap V(P)$ as being the negative of this class.

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Thanks alot, once again! I was wondering, did you mean Bertini's Theorem instead of Bezout? – FMN Jul 5 '12 at 17:56
@FMN: Dear FMN, Yes, thanks for pointing out this misstatement! Regards, – Matt E Jul 5 '12 at 20:53

This is more of a comment than an answer (but I cannot comment yet). Isn't it the case that "the" canonical divisor is any divisor representing the linear equivalence class of the canonical bundle? We would get intersection points only after choosing a representative (and coordinates), and the points may depend on our choice. Only the intersection class (in a Chow ring) should be intrinsic, I think. I would appreciate hearing more on this too!

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As far as I know, there are three types of objects to consider; sections of the canonical sheaf, canonical divisors, and canonical divisor classes. I think that you're right that often times people say "the canonical divisor" to mean the divisor class, but I think FMN's question is about the specific divisor, and not its class. – jrajchgot Jul 1 '12 at 23:40
@jrajchgot: Dear jrajchgot, There is no "specific divisor"; no one member of the canonical divisor class is any more specific than any other. What's worse (from the point of view of making things concrete), in the particular case of projective space, there are no effective members of this divisor class at all. Regards, – Matt E Jul 5 '12 at 3:15
Dear Matt, What term would you use to refer to a specific representative of the class (in order to avoid ambiguity)? I have, until now, been using "canonical divisor" (vs. "divisor class") but perhaps this is dangerous.. – jrajchgot Jul 5 '12 at 18:23
@jrajchgot: Dear jrajchgot, You could say "a canonical divisor" (with "a" instead of "the" to indicate that it is just one of many possible choices), but to be safe I might write something like "let $D$ be a representative of the canonical divisor class" if I really needed a specific choice. On the other hand, often people write "let $K_X$ be the canonical divisor" and treat $K_X$ as a specific divisor rather than a class of such (e.g. in Riemann--Roch calculations), but this is fine as long as everything that follows is invariant with respect to replacing a divisor by a linear equivalent ... – Matt E Jul 5 '12 at 20:52
... one (but only if this invariance holds!). Regards, – Matt E Jul 5 '12 at 20:53

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