# Direct Consequence of Seesaw Theorem

I am reading Hideaki Ikoma & Atsushi Moriwaki & Shu Kawaguchi「The Mordell-Faltings theorem」. The following is a quotation from this book. (lemma 2.19)

Let $$F$$ be a field of characteristic zero and let $$X$$ and $$T$$ be geometrically irreducible algebraic varieties over $$F$$. We assume that $$X$$ is projective over $$F$$. Let $$p \colon X \times T \to T$$ be a projection.

Suppose that $$L_1$$ and $$L_2$$ are line bundles on $$X \times T$$. The seesaw theorem tells us that $$L_1 \otimes L_2^{-1} \in p^* (Pic(T))$$ if and only if $$L_1|_{X \times \{ t \}} \cong L_2|_{X \times \{ t \}}$$ for any $$t \in T$$. Further, if there is an $$x \in X(F)$$ with $$L_1|_{ \{ x \} \times T} \cong L_2|_{ \{ x \} \times T}$$ , then $$L_1$$ and $$L_2$$ are isomorphic. Indeed, we have $$L_1 \otimes L_2^{-1} \cong p^* M$$ for some line bundle $$M$$ on $$T$$, and the restriction $$p^*M|_{\{ x \} \times T}$$ is trivial, so $$M$$ is trivial.

My question:

(1) I think the condition $$\forall t \in T \quad L_1|_{X \times \{ t \}} \cong L_2|_{X \times \{ t \}}$$ implies $$L_1 \cong L_2$$. The following is my proof.

Let $$\phi_t \colon L_1|_{X \times \{ t \}} \to L_2|_{X \times \{ t \}}$$ be an isomorphism. Then $$\phi_t$$ induces an isomorphism $$(\forall p \in X \times T) \; (L_1)_p / \mathfrak{m_p} (L_1)_p \to (L_2)_p / \mathfrak{m_p} (L_2)_p$$. Since $$X \times T$$ is integral and $$L_1$$ and $$L_2$$ are line bundle, $$L_1$$ and $$L_2$$ are isomorphic.

Why is the condition "$$\exists x \in X(F) \; L_1|_{ \{ x \} \times T} \cong L_2|_{ \{ x \} \times T}$$" necessary?

(2) How to show "M is trivial" from "$$p^*M|_{\{ x \} \times T}$$ is trivial".

If you pull back a line bundle $$M$$ from $$T$$, then it's restriction to $$X\times\{t\}$$ will be trivial for all $$t$$. If what you claim were true, then this would show that $$p^*M \cong \mathscr O_{X\times T}$$. Of course, there is no reason this should be true in general.
For instance, take $$X=T = \mathbb P^1$$ and $$M$$ any non trivial line bundle.
I might be missing something but your proof definitely does not even get off the ground. Consider that any two line bundles have isomorphic fibers at all points. The key point here is to construct a global map $$\phi: L_1 \to L_2$$ whose restriction to each point is an isomorphism. We are given maps $$\phi_t$$ but no way to glue them together.
• Show that $p^*M|_{\{x\}\times T} = M$. This is just because pulling back to $X\times T$ and then $\{x\}\times T$ is the same as pulling back to $\{x\}\times T$ directly but this last map is just the identity. – Asvin Jun 25 at 2:40