TITLE: Principal $G$-bundles in Zariski vs étale topology
QUESTION [22 upvotes]: 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 étale 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 étale topology but not locally trivial in the Zariski topology? You are allowed to pick your group $G$. The groups $GL(n)$ and $SL(n)$ are "special" and so étale locally trivial implies Zariski locally trivial for those choices of $G$.

REPLY [10 votes]: 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.