There is a very basic theorem for the Zariski topology.


Let X = Spec(R) and Y=Spec(R/I) for I some reduced ideal. Y obtains a topology two ways, one is the subspace topology as a subset of X and another as the spectrum of a ring. These topologies are the same by the correspondence between ideals of R containing I and ideals in R/I.


Is there a close statement to this in the etale toplogy? There are two natural ways to understand open sets on Y, those which come from etale neighborhoods of X base changed to Y and those which are etale neighborhoods of Y.


I did a computation today in a very special case and it seems that both of these topologies seem to be 'the same'.


Does anyone know if this statement is true in a general context and where I might locate this resource?


Thanks.



The etale site of a closed subscheme and its etale Grothendieck subtopology 
The statement is at least true Zariski locally. That is, given an étale map $V\to Y$, there exists a Zariski open cover $X=\bigcup X\_i$ so that the pullback of $V$ to $Y\cap X\_i$ is the restriction of an étale neighborhood of $X\_i$.


To see this, use the structure theorem for étale morphisms: [Theorem 34.11.3](http://www.math.columbia.edu/algebraic_geometry/stacks-git/locate.php?tag=025C) in the chapter on étale morphisms in the [Stacks Project](http://www.math.columbia.edu/algebraic_geometry/stacks-git/index.html).



It says that any étale morphism to $Y=Spec(R/I)$ is Zariski locally† an open subscheme $V$ of $Spec((R/I)[t]\_{\bar f'}/(\bar f))$, where $\bar f\in (R/I)[t]$ is monic. Let $f\in R[t]$ be an arbitrary monic lift of $\bar f$. Then since the Zariski topoplogy of $Spec((R/I)[t]\_{\bar f'}/(\bar f))$ is the restriction of the Zariski topology of $Spec(R[t]\_{f'}/(f))$, there is some open subscheme $U$ of $Spec(R[t]\_{f'}/(f))$ so that $V = U\cap Spec((R/I)[t]\_{\bar f'}/(\bar f))$. This $U$ is étale over $X$ and pulls back to $V$. 



† I'm implicitly replacing $X=Spec(R)$ and $Y=Spec(R/I)$ by localizations $Spec(R\_g)$ and $Spec(R\_g/I\cdot R\_g)$ in the rest of the argument. The open cover $X=\bigcup X\_i$ consists of these $Spec(R\_g)$'s and the complement of $Y$.


