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Let $X$ be a projective scheme (not necessarily reduced). Under what assumption is it true that $h^0(\mathcal{O}_X)=1$? In general can we bound $h^0(\mathcal{O}_X)$ by the dimension of the projective space in which it is emebedded? (Here $h^0(\mathcal{O}_X):=\dim H^0(\mathcal{O}_X)$)

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No. This has much less to do with dimension than it does with the number of components. –  Matt Nov 6 '13 at 18:47
@Matt: How is it related to the number of components? –  Chen Nov 6 '13 at 19:21
For example, for a smooth and projective scheme over a field, it equals the number of components. Just try out $X=\prod_1^n Spec(k)$. Note this fits inside a small dimensional projective space. –  Matt Nov 6 '13 at 20:17
@Matt: Does a similar result hold for projective but not reduced? –  Chen Nov 6 '13 at 20:18

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