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Recently I had to compute the rational singular cohomology ring of a hypersurface in a product of projective spaces. I managed to do it, but only by interpreting this variety as a blowup of projective space (so I wasn't using the fact that it was a hypersurface in a toric variety).

My question is:

Given a nonsingular projective toric variety $X$ and a line bundle $L \to X$, is there a simple way of writing down the singular cohomology ring of the hypersurface cut out by $L$?

By "simple" I am thinking of something not much more involved than the presentation given in Fulton's book of the integralrational cohomology ring of a nonsingular projective varietysymplectic toric manifold in terms of the Stanley-Reisner ideal. I would be happy to know the answer in the special case that $X$ and the hypersurface are both Fano.

The original question that came to my mind was whether there's aan easy way of computing cohomology for a nonsingular complete intersection in a nonsingular projective toric variety, but after doing some browsing around I would guess that this is too much to hope for.

Recently I had to compute the rational singular cohomology ring of a hypersurface in a product of projective spaces. I managed to do it, but only by interpreting this variety as a blowup of projective space.

My question is:

Given a nonsingular projective toric variety $X$ and a line bundle $L \to X$, is there a simple way of writing down the singular cohomology ring of the hypersurface cut out by $L$?

By "simple" I am thinking of something not much more involved than the presentation given in Fulton's book of the integral cohomology ring of a nonsingular projective variety. I would be happy to know the answer in the special case that $X$ and the hypersurface are both Fano.

The original question that came to my mind was whether there's a way of computing cohomology for a nonsingular complete intersection in a nonsingular projective toric variety, but after doing some browsing around I would guess that this is too much to hope for.

Recently I had to compute the rational singular cohomology ring of a hypersurface in a product of projective spaces. I managed to do it, but only by interpreting this variety as a blowup of projective space (so I wasn't using the fact that it was a hypersurface in a toric variety).

My question is:

Given a nonsingular projective toric variety $X$ and a line bundle $L \to X$, is there a simple way of writing down the singular cohomology ring of the hypersurface cut out by $L$?

By "simple" I am thinking of something not much more involved than the presentation of the rational cohomology ring of a symplectic toric manifold in terms of the Stanley-Reisner ideal. I would be happy to know the answer in the special case that $X$ and the hypersurface are both Fano.

The original question that came to my mind was whether there's an easy way of computing cohomology for a nonsingular complete intersection in a nonsingular projective toric variety, but after doing some browsing around I would guess that this is too much to hope for.

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Is there an easy way to write down the singular cohomology of a hypersurface in a toric variety?

Recently I had to compute the rational singular cohomology ring of a hypersurface in a product of projective spaces. I managed to do it, but only by interpreting this variety as a blowup of projective space.

My question is:

Given a nonsingular projective toric variety $X$ and a line bundle $L \to X$, is there a simple way of writing down the singular cohomology ring of the hypersurface cut out by $L$?

By "simple" I am thinking of something not much more involved than the presentation given in Fulton's book of the integral cohomology ring of a nonsingular projective variety. I would be happy to know the answer in the special case that $X$ and the hypersurface are both Fano.

The original question that came to my mind was whether there's a way of computing cohomology for a nonsingular complete intersection in a nonsingular projective toric variety, but after doing some browsing around I would guess that this is too much to hope for.