When is the double cover of a smooth curve reducible Let $C$ be a smooth projective plane curve. Suppose $f:C'\rightarrow C$ is a double cover of $C$ ramified at $d$ points. Suppose $C'$ is irreducible, then the Riemann Hurwitz formula says that $d$ has to be even, is that correct?
Consequently, if I know that $d$ is odd, does that mean that the double cover $C'$ is reducible?
 A: Yes, I think you are correct. The RH-formula for a branched double cover $f:S' \rightarrow S$ gives:$$ \chi(S') = 2\chi(S) - \sum_{p \in S'}(E_{p}-1) $$ where $e(p)$ is the ramification index at $p\in S'$. Since $f$ is a ramified cover of degree 2 the map is $2:1$ on the complement of the ramification subset, implying that $e(p) \leq 2$ for all $p \in S'$, since otherwise we get a point in $S$ (over which the map is regular) whose pre-image has more than $3$ preimages: a contradiction. 
Hence, $\sum_{p \in S'}(E_{p}-1)$ is equal to the number of ramification points in $S'$, and is clearly even by the formula.
The definition of ramified cover implies easily that a ramified cover of a Riemann surface has to be a (possibly disconnected) Riemann surface. So, if a curve is branch covered by some reducible curve, then the irreducible components are pairwise disjoint smooth curves. Then we can apply the above argument to each component of the domain, giving in general that the number of ramification points is always even. Although I should mention that if $S'$ has more than one component, then we have that $S'$ is just two copies of $C$, in particular it is an unramified cover.
