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I am currently taking my first course in algebraic geometry and am stuck at te following problem, which I am sure is simple. Consider $Y := \mathbb P^1 \times \{x\}$ as a closed subscheme of $\mathbb P^1 \times \mathbb P^1$. Suppose you have a line bundle $\mathcal L$ on $\mathbb P^1 \times \mathbb P^1$, which is of the form $p_1^*(\mathcal O(n_1)) \otimes p_2^*(\mathcal O(n_2))$, where $p_1,p_2$ are the projections onto the two factors. What is $\mathcal L \otimes \mathcal O_Y$? I am not able to rigorously prove it.

Anonymous
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    Dear anonymous, note that $L \otimes O_Y$ is not a locally free sheaf on $\mathbf P^1 \times \mathbf P^1$: it has nonzero stalks at the points of $Y$, but zero stalks elsewhere. By contrast, $p_1^* O(n_1)$ is the pullback of a line bundle (=locally free sheaf of rank 1), hence is a line bundle. So they can't be the same. –  Feb 13 '14 at 12:39
  • @AsalBeagDubh: Thank you very much for pointing this out. I have edited the question now. Is it possible to explicitly determine what $\mathcal L \otimes \mathcal O_Y$ is? Is it $i_*(\mathcal O(n_1))$, where $i : Y \to \mathbb P^1 \times \mathbb P^1$ is the inclusion? – Anonymous Feb 13 '14 at 13:50
  • Yes, that's exactly right. –  Feb 13 '14 at 13:51
  • This motivates me to ask another question: Let $\mathcal F$ be any sheaf on $\mathbb P^1 \times \mathbb P^1$. Is it true that $\mathcal F \otimes \mathcal O_Y$ is always of the form $i_*\mathcal G$, for some sheaf $\mathcal G$ on $Y$? – Anonymous Feb 13 '14 at 15:05
  • Yes, $\mathcal{F} \otimes \mathcal{O}Y$ is supported only on $Y$ (tensoring with $\mathcal{O}_Y$ means restricting to the subscheme $Y$), so you can think of it as $i(i^ \mathcal{F})$. – Jake Levinson Feb 14 '14 at 03:49
  • Someone should write an answer to this. Perhaps @AsalBeagDubh :) – Alex Youcis Feb 14 '14 at 06:44

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(I'll post the comments as a CW answer to remove this from the unanswered pile. Other users should feel free to improve it.)

For your original question: if $L = p_1^* \mathcal{O}(n_1) \otimes p_2^* \mathcal{O}(n_2)$, then $L \otimes \mathcal{O}_Y$ is indeed isomorphic to $i_* \mathcal{O}(n_1)$.

The question in the comments was answered by Jake Levinson: if $F$ is a sheaf (of modules) on $X$, and $Y \subset X$ a closed subscheme, then $F \otimes \mathcal{O}_Y$ is isomorphic to $i_*(i^* F)$.