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Let $k$ be an algebraically closed field of characteristic 0.

The seesaw principle in algebraic geometry usually goes like this: let $T$ be a complete variety, let $X$ be an integral scheme of finite type over $k$, and let $L$ be a line bundle on $T \times X$. Then $L = 0$ if and only if $L|_{T \times \{x\}} = 0$ for all $x \in X$ and $L|_{\{t\} \times X} = 0$ for some $t \in T$.

My problem: different sources have different definitions for the notion of "variety" $-$ some insist that a variety should be (geometrically) integral, some just require reducedness, but not connectedness; all require separatedness and of finite type over $k$. These different sources state the seesaw principle in the same way. It is not clear to me, however, whether the (geometrically) connectedness of $T$ is used anywhere in the proof of the seesaw principle.

Question: Does the seesaw principle hold if $T$ is a complete, reduced, separated scheme of finite type over $k$ (but not necessarily (geometrically) connected)? If not, where does the proof break?

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  • $\begingroup$ Never heard of this, but I bet you need some kind of connectedness. Otherwise, let $T=T_1\cup T_2$ be disconnected, and take $L_1$ a nonzero line bundle on $T_1\times X$ such that $L_1|_{T_1\times\{x\}}=0$ for all $x\in X$. I suppose this exists since the seesaw principle would be phrased otherwise if it weren't. Pick the zero line bundle $L_2$ on $T_2 \times X$. This gives you a line bundle $L$ on $T\times X$ which is nonzero, but it can't be by the seesaw principle (picking a point $t\in T_2$). $\endgroup$ – Jesko Hüttenhain Jun 7 '16 at 13:44
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    $\begingroup$ Why not just write down the proof and see where (if at all) it breaks? Have you tried that? $\endgroup$ – Mohan Jun 7 '16 at 13:58

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