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if a subvariety of codimension n is given by an ideal of polynomials with n generators, then the homology class of the variety is given by the intersection product of the classes of the individual generators' zero set. In this case, all the (zero set of the) individual generators intersect transversally.

The example I have in mind is something like the zero locus of $ \langle z_0, z_1 \rangle$ inside $\mathbb{P}^3$, where the z_i are the usual projective coordinates. Then the class of this zero locus should be just $[z_0] \cdot [z_1] = H^2$, where H is the hyperplane class.

However, what is the situation if the ideal has more than n generators? An example would be something like $\langle a \, d - b \, c , \, a\,x + b\,y , \, c\,x + d\,y \rangle$ in a space where $a,b,c,d,x,y$ are sections of some line bundles. How can I express the class of this variety in terms of the classes of the sections?

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    $\begingroup$ There is no reason that you can express that class in terms of the classes of generators of the ideal. For instance, consider any Grassmannian that is not a projective space. Since the cohomology ring, as a ring, is not generated by (Poincare duals of) divisor classes, it is impossible to express all cohomology classes as polynomials in these divisor classes. If your subvariety is determinantal (which your specific equations suggests), and with appropriate transversality hypotheses, you can use Thom-Porteous. $\endgroup$ Dec 3, 2014 at 13:54
  • $\begingroup$ To the first point: You are right, I need to point out that I work in a toric variety, so the cohomology ring should be generated by the toric divisors? Regarding the second suggestions: I am afraid I have no clue about determinantal varieties. From a quick glance on Wikipedia, I don't think this is the general setting of my problem, even though my example may seem to be falling into that category. $\endgroup$
    – moep
    Dec 3, 2014 at 14:51
  • $\begingroup$ Can you maybe just illustrate the Thom-Porteous formula for this example, assuming the appropriate requirements? $\endgroup$
    – moep
    Dec 3, 2014 at 15:12
  • $\begingroup$ "Can you maybe just illustrate ..." Consider the $2\times 3$ matrix whose first row $3$-vector is $(a,c,-y)$ and whose second row $3$-vector is $(b,d,x)$. The locus where this matrix has rank $1$ is precisely your locus. Such a locus is called "determinantal". $\endgroup$ Dec 3, 2014 at 16:42
  • $\begingroup$ Thanks for that last hint! After some digging and experimenting, I think I got an ideal how it works! And it produces the correct result for the known case of the Segre embedding of $\mathbb{P}^2 \times \mathbb{P}^1$ into $\mathbb{P}^5$, which the example corresponds to in the case that a,b,...,y are homogenous coordinates of $\mathbb{P}^5$. $\endgroup$
    – moep
    Dec 5, 2014 at 13:28

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