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Let $\mathcal{L}$ be a line bundle over a proper variety $X/k$. Choose a $k$-rational point $P$ in some fibre of $\mathcal{L}$.

Why are there no non-trivial automorphisms of $\mathcal{L}$ fixing $P$?

Does this have something to do with $\Gamma(X,\mathcal{O}_X) = k$? ($\mathrm{Hom}(\mathcal{L},\mathcal{L}) = \mathrm{Hom}(\mathcal{O}_X,\mathcal{O}_X) = \Gamma(X,\mathcal{O}_X) = k$)

user5262
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1 Answers1

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The sheaf of homomorphisms $\mathcal L\to \mathcal L$ is isomorphic to the structural sheaf: $$\mathcal {\text { Hom}}(\mathcal L, \mathcal L)\cong \mathcal O$$ so that taking global sections we have the isomorphism $$ Hom(\mathcal L, \mathcal L)\cong \mathcal O( X)$$ If $X$ is complete and integral we deduce $$ Hom(\mathcal L, \mathcal L)\cong \mathcal O( X)=k$$ and $$\text {Isom }(\mathcal L, \mathcal L)\cong \mathcal O^*( X)=k^*$$ from which your result follows.

Georges Elencwajg
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  • Thank you. I am wondering if my claim is true at all for $P = 0 \in \mathcal{L}_x(k)$? – user5262 Oct 03 '13 at 08:30
  • No. The essence of the result is that an automorphism of a line over a field is the identity as soon as it fixes a non-zero vector. You should modify your question accordingly. – Georges Elencwajg Oct 03 '13 at 08:33