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 $G$.
Algebraic Geometry – Principal G-bundles in Zariski vs Étale Topology
algebraic-geometryprincipal-bundleszariski-topology
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Let $\pi : P \rightarrow X$ be a $GL(n)-$torsor and is locally trivial in the etale topology. We want to prove that it is locally trivial in the Zariski topology.Hence forth we denote $GL(n)$ by $G$ for convenience.
First let us construct the natural associated vector bundle. We just imitate the classical construction. Let $g \in GL(n)$ act on $P \times \mathbb{A}^n$ by $g.(x,y) = (x.g, g^{-1}.y)$, where $GL(n)$ acts on the right on $P$ and in a natural manner from the left on $\mathbb{A}^n$. Note that this action is free since the action is free on $P$. Let us look at the $GL(n)$ orbit of the action.
Claim : All GL(n) orbit on $P$ is contained in an open affine subset of $P$.
Proof of Claim : We know that for $p \in P$, we have $p.G = \pi^{-1}(\pi(p))$. Also note that $\pi$ is an affine map, since it is affine after etale base change. This is a statement that "affine morphism is local on the target". Now choose an open affine neighbourhood of $\pi(p)$, say $U_{\pi(p)}$ and let $U_p := \pi^{-1}(U_{\pi(p)})$. Since $\pi$ is affine, hence $U_p$ is affine and it clearly contains the orbit. Hence the claim.
Using the claim, we get that orbit of $GL(n)$ on $P \times \mathbb{A}^n$ is contained in an open affine subset since $\mathbb{A}^n$ is affine. Also note that the action is free. This allows us to form a quotient space say $E$ which has an obvious map to $X$ which comes from quotient of the $G-$ equivariant projection map $P \times \mathbb{A}^n \rightarrow P$.
Since $GL(n)$ is a smooth group scheme, $P$ is smooth over $X$. This follows from the following : Let $U \rightarrow X$ be etale cover such that $P \times_X U \rightarrow U$ is locally trivial. Since $GL(n)$ is smooth scheme, hence this is a smooth map. Thus we have the following situation $P \times_X U \rightarrow P$ is a smooth map and $P\times_X U \rightarrow U \rightarrow X$ is a smooth map, hence the map $P \rightarrow X$ is also smooth. This statement is known as "smoothness is etale local on the target"
It can be checked from the construction that $E$ is also etale locally trivial with fibers $\mathbb{A}^n$ and hence $E \rightarrow X$ is smooth affine. Let us assign a name $f : E \rightarrow X$.
Let $U_i \xrightarrow{\phi_i} X$ be etale cover such that for all $i$, we have $E \times_X U_i \rightarrow U_i$ is trivial. Thus we have $\phi^*(f_*\mathcal{O}_E) \cong \mathcal{O}_{U_i}[T_1,\dots T_n]$. Let $F_i = \oplus \mathcal{O}_{U_i}T_i$. Note that since $E$ is locally trivial for etale topology, we automatically have a descent data for $\lbrace F_i, \lbrace{U_i\phi_i} \rbrace \rbrace$(I have supressed the notation for coordinate transformations). Thus we have a zariski locally free sheaf $F$ on $X$, such that $\phi_i^*F \cong F_i$. We have $Spec(Sym(F_i)) \cong E \times_X U_i = \phi_i^*(E) \cong \phi_i^*(f_*\mathcal{O}_E)$. This implies that $Sym(F_i) \cong \phi_i^*(f_*\mathcal{O}_E)$. Thus we have a morphism of (effective)descent data and hence we have a map, infact an isomorphism $E \cong Spec(SymF)$(see 3).
This shows that $E$ is infact locally trivial in the Zariski topology. Now the rest should be clear from the answer here : https://mathoverflow.net/a/168004/58056
I will write it here for completeness. We have $P \cong \underline{Isom}(\mathbb{A}^n_S, E)$. Since $E$ is Zariksi locally trivial, we obtain that $P$ is locally trivial.
Here are some references for the descent arguments.
https://stacks.math.columbia.edu/tag/02L5 a lemma which says that the property of morphism being affine is local on the base for the fppf topology and hence also in the etale topology.
https://stacks.math.columbia.edu/tag/023B is the definition for the definition of descent and morphism of descent data for quasi-coherent sheaves.
https://stacks.math.columbia.edu/tag/023E says that the descent data is always effective and also implies that morphism of descent data gives a unique morphism for the quasi-coherent sheaves.
There might be some gaps in the argument. I do not know of a way to avoid all this terminology except maybe by following the line of argument given in the comment above(http://www-personal.umich.edu/~takumim/takumim_Spr14Thesis.pdf).
This follows from the infinitesimal lifting criterion for smoothness: An algebraic stack $\mathcal{X}$ locally of finite type is smooth if and only if for every nilpotent thickening $S \to S'$ of affine schemes, any map $S \to \mathcal{X}$ extends to $S'$. If $G$ is an algebraic group, its classifying space $BG := [pt/G]$ is an algebraic stack locally of finite type for the etale topology. This is so simply because $BG$ has a smooth cover by a point.
I'm not sure what the best reference is for the infinitesimal lifting criterion, but you can see one direction of the proof on page $36$ here.
Best Answer
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.