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Let $G$ be an (affine) algebraic group over say $\mathbb{C}$. A principal $G$-bundle is a scheme $P$ with a $G$ action and a $G$-invariant morphism of schemes $\pi:P \to X$ that is etale locally on $X$ isomorphic to the trivial $G$-bundle $U \times G \to U$. What is a simple example of a $G$-bundle that is locally trivial in the etale topology but not locally trivial in the Zariski topology? You are allowed to pick your group $G$. The groups $GL(n), SL(n)$ are "special" and so etale locally trivial implies Zariski locally trivial for those choices of $G$.

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An excellent question, alex! And, please, don't forget to upvote and accept the one that answers it to your satisfaction. – Georges Elencwajg Apr 2 '12 at 12:22
up vote 5 down vote accepted

Let $X=Spec ~~\mathbb C [X,X^{-1}]$ and let $Y=Spec ~~\mathbb C [X,X^{-1},Y]/(Y^2-X)$, with $f:Y \rightarrow X$ the natural map. Here $Y$ is a degree 2 covering of $X$ with a $\mathbb Z /2 \mathbb Z$ action taking $y$ to $-y$. This is the punctured affine line wound twice around itself.

In the complex (and hence etale) topology, this action makes $Y \rightarrow X$ into a principal $\mathbb Z /2 \mathbb Z$-bundle, but it's not the case in the Zariski topology. A Zariski open $U$ is the complement of a finite point set of $X$, so the restriction of $Y$ to $U$ is also the complement of a finite point set of $Y$. But if $Y$ were locally trivial then this means that $Y_{|U}=U \bigsqcup U$ with each $U$ open in $Y$. This contradicts the fact that each $U$ must be infinite.

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