Suppose $X$ and $Y$ are two smooth surfaces (over the complex numbers), and $f: X \to Y$ is a finite flat morphism of degree two. Is it necessarily true that the locus where $f$ is not a smooth morphism (i.e. the ramification locus) always one dimensional (if not empty)?
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1$\begingroup$ Yes. The result which is true under more general conditions, is called "purity of the branch locus". I don't have access to my books right now, otherwise I'd give you a reference. $\endgroup$– Donu ArapuraCommented Jun 1, 2013 at 4:02
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2$\begingroup$ yes, in this casethe locus is defined by one equation, the determinant. $\endgroup$– roy smithCommented Jun 1, 2013 at 5:11
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1$\begingroup$ Since $X$ and $Y$ are smooth surfaces (this works for smooth varieties of the same dimension), the locus where $f$ is not a smooth morphism is given by the vanishing of the Jacobian determinant (as said by Roy), say, written in local coordinates. Purity of the branch locus (Nagata; see, eg, Grothendieck's SGA 2, X, 3.4), is more general, because it extends to schemes (if $f\colon X\to Y$ is a quasi-finite morphism of integral schemes, $X$ normal and $Y$ regular, and if $f$ is unramified outside a 2-codimensional subset, then $f$ is étale) and does not assume that $X$ is itself regular. $\endgroup$– ACLCommented Jun 1, 2013 at 8:54
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Yes. Otherwise there would be an isolated ramification point in $Y$. A link of that point is $S^3$, which does not have a nontrivial double cover. So upstairs, $X$ looks near that point like two ${\mathbb C}^2$s glued at a point, which isn't smooth.
answered Jun 1, 2013 at 2:02
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$\begingroup$ I feel a little silly writing this topological answer, given the beautiful algebraic geometry answers above, but I'll add: note how things are different if $Y$ were codim $1$, since the link would now be $S^1$, which does indeed have a nontrivial double cover. $\endgroup$ Commented Jun 4, 2013 at 2:20