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I have a stupid question: assume $X \subset P:=\mathbb{P}^{N_1}\times \mathbb{P}^{N_2} $ closed, irreducible, Cohen-Macaulay, not a product of two varieties and non degenerate, meaning that it is not contained in any hyperplane $H\times K$ where $H$ is an hyperplane in the first factor and $K$ an hyperplane in the second. Is there a lower bound depending on $N_1$ and $N_2$ for the degree of $X$ in terms of the dimension of $P$?

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What do you mean by hyperplane of that product? –  Mariano Suárez-Alvarez May 7 '12 at 18:18
    
@Mariano Suárez Alvarez Sorry I edited –  unki May 7 '12 at 18:55
    
If $H$ and $K$ are hyperplanes, then $H \times K$ is not a hyperplane in the Segre embedding of $P^{N_1} \times P^{N_2}$, nor is it necessarily contained in one. –  Michael Joyce May 7 '12 at 19:22

1 Answer 1

No - there are plane curves of arbitrarily high degree. (Take $N_1 = 2, N_2 = 0$.)

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you are right, but a line is degenerate, in $\mathbb{P}^3$ every line is also degenerate and also each conic is degenerate so the lower bound is 3 –  unki May 7 '12 at 18:58
    
You need an assumption on $\dim X$ and not just $\dim P$. There are non-degenerate conic hypersurfaces of every dimension. –  Michael Joyce May 7 '12 at 19:23

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