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Let $X$ be a projective variety with ample sheaf $\mathcal{O}_X(1)$. Then $H^*(\oplus_n \mathcal{O}_X(n))$ is a graded algebra via the cup product: $H^i(\mathcal{O}(n)) \otimes H^j(\mathcal{O}(m)) \to H^{i+j}(\mathcal{O}(n) \otimes \mathcal{O}(m)) \cong H^{i+j}(\mathcal{O}(n+m))$. Remark that each individual homogeneous component is a graded abelian group (via $n$). Does this "bigraded" cohomology ring have a name, and is it studied in the literature?

If $X,Y$ are projective varieties such that the corresponding "bigraded" cohomology rings are isomorphic, do we then have $X \cong Y$? If this is false, what about the special case $Y=\mathbb{P}^d$ (here the cohomology ring is quite simple).

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    @MartinBrandenburg: without "@QiL", I was not aware of your question.2012-11-03

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If I understand your construction correctly, the bi-graded algebra is $ \oplus_{i,j} H^i(X, O_X(j))$ and the component of degree $(i,j)$ is $H^i(X, O_X(j))$. So if an isomorphism $ \oplus_{i,j} H^i(X, O_X(j)) \simeq \oplus_{i,j} H^i(Y, O_Y(j))$ of bi-graded algebras is required to preserve bi-degrees, hence induces linear isomorphisms $ H^0(X, O_X(j))\simeq H^0(Y, O_Y(j))$ then $X\simeq \mathrm{Proj}\left(\oplus_{j\ge 0} H^0(X, O_X(j))\right) \simeq \mathrm{Proj}\left(\oplus_{j\ge 0} H^0(Y, O_Y(j))\right) \simeq Y.$

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    - Thanks a lot!2012-11-07