Lemma 31.17.1. Let $\pi : X \to Y$ be a finite morphism of schemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $y \in Y$. There exists an open neighbourhood $V \subset Y$ of $y$ such that $\mathcal{L}|_{\pi ^{-1}(V)}$ is trivial.

Proof. Clearly we may assume $Y$ and hence $X$ affine. Since $\pi$ is finite the fibre $\pi ^{-1}(\{ y\} )$ over $y$ is finite. Since $X$ is affine, we can pick $s \in \Gamma (X, \mathcal{L})$ not vanishing in any point of $\pi ^{-1}(\{ y\} )$. This follows from Properties, Lemma 28.29.7 but we also give a direct argument. Namely, we can pick a finite set $E \subset X$ of closed points such that every $x \in \pi ^{-1}(\{ y\} )$ specializes to some point of $E$. For $x \in E$ denote $i_ x : x \to X$ the closed immersion. Then $\mathcal{L} \to \bigoplus _{x \in E} i_{x, *}i_ x^*\mathcal{L}$ is a surjective map of quasi-coherent $\mathcal{O}_ X$-modules, and hence the map

$\Gamma (X, \mathcal{L}) \to \bigoplus \nolimits _{x \in E} \mathcal{L}_ x/\mathfrak m_ x\mathcal{L}_ x$

is surjective (as taking global sections is an exact functor on the category of quasi-coherent $\mathcal{O}_ X$-modules, see Schemes, Lemma 26.7.5). Thus we can find an $s \in \Gamma (X, \mathcal{L})$ not vanishing at any point specializing to a point of $E$. Then $X_ s \subset X$ is an open neighbourhood of $\pi ^{-1}(\{ y\} )$. Since $\pi$ is finite, hence closed, we conclude that there is an open neighbourhood $V \subset Y$ of $y$ whose inverse image is contained in $X_ s$ as desired. $\square$

Comment #7180 by DatPham on

Is there any chance that this Lemma holds true for a general vector bundle (rather than just line bundles)?

Comment #7181 by DatPham on

I was just wondering if any vector bundle on $X$ can be trivialized by an affine open cover coming from $Y$.

Comment #7182 by on

Yes, this is true for finite locally free modules of constant rank $r$. Basically, the argument given in the proof works, except instead of finding a single section you choose $s_1, \ldots, s_r$ which map to a basis of $\mathcal{L}_x/\mathfrak m_x\mathcal{L}_x$ for all $x \in E$.

Comment #7183 by DatPham on

Ah, I see. Thanks for your answer. It seems to me that this argument is essentially the one used to show that any vector bundle of constant fiber rank over a semi-local ring is in fact trivial.

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