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Consider the plane nodal cubic $X$ given by the equation $y^2=x^2(x+1)$ over a field $k.$ It is not too hard to show that $Y$ has a finite étale covering $X$ of degree $2.$ One does this in the following manner. Take two copies of the normalization $\pi:\tilde{X} \rightarrow X,$ call them $\tilde{X_1}$ and $\tilde{X_2}$ respectively. The fiber $\pi^{-1}(0)$ of the singularity consists of two points $x_1$ and $x_2.$ Call the corresponding points in $\tilde{X_1}$ by $y_1,y_2$ and the corresponding points in $\tilde{X_2}$ by $z_1,z_2.$ We then glue $\tilde{X_1}$ and $\tilde{X_2}$ along these points, so that $y_1$ is glued along $z_2$ and $y_2$ is glued along $z_1.$ One shows that this is étale without too much pain.

I'm curious, if we instead glue $y_1$ along $z_1$ and $y_2$ along $z_2,$ how would this change the result? Would we still get a non-trivial étale covering?

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    $\begingroup$ Abstractly $\tilde X$ is the affine line. There is an automorphism which interchanges any two points, so in particular $x_1, x_2$. So you should get the same covering up to isomorphism. $\endgroup$ Feb 28, 2016 at 15:53
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    $\begingroup$ Yes but the exchanges is not compatible with the morphism $\tilde{X}\to X$. $\endgroup$ Feb 28, 2016 at 20:53

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No, the corresponding map is not étale. The morphism $\tilde{X}\to X$ is basically the identification of the two points $x_1$ and $x_2$. Around the singularity of $X$, there are thus two "branches", corresponding to the image of the neighbourhood of $x_1$ and $x_2$.

If you glue $y_1$ with $z_1$, then a neighbourhood of the point obtained will be send only on one branch, and will be locally a $2:1$-map. The map is thus not étale. It should be possible to do it in affine coordinates if you want.

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