Smooth branch divisor implies smooth covering space

Question

Let $c: S'\to S$ be a branched cover of a smooth projective surface $S$ that is branched over the smooth divisor $B\subset S$. Why does smoothness of $S$ and $B$ imply smoothness of $S'$?

Answer 1

Depending on the tools at your disposal, this is a nice exercise. I think the most basic way is to use the Jacobian criterion:

Since smoothness is local, we can suppose that we are looking at the double cover $\mathrm{Spec}(A[t]/(t^2-f))\to \mathrm{Spec}(A)$ for some $f\in A = k[x,y]$. That is, locally, $S$ looks like $\mathrm{Spec}(k[x,y,t]/(t^2-f))$. To see that this is smooth, we only have to show that the Jacobian is nowhere trivial. More precisely, the singular locus is given by the ideal $(2t,\partial f/\partial x,\partial f/\partial y)$ and for this to be the unit ideal, we have to have that $(\partial f/\partial x,\partial f/\partial y)$ is the unit ideal in $k[x,y]$, which is the same as smoothness of $f$.