TITLE: Quadratic twist of curve defined over finite field
QUESTION [10 upvotes]: I was reading the paper on "curves of every genus with many points II" by: Elkies, Howe, et al. And some of the terms are not clear to me. Is there any elaborate exposition on these stuffs?
In particular I appreciate an explanation on the following. Let $C$ be a smooth, projective curve over field $\mathbb{F}_q$ ($q$ is a power of prime $p$) and $B\rightarrow C$ a degree-2 cover. Let $B'$ be quadratic twist of $B$. So is there any geometric intuition behind quadratic twist of a curve and why do we have $\#B(\mathbb{F}_q)+\#B'(\mathbb{F}_q)=2\#C(\mathbb{F}_q)$?

REPLY [3 votes]: I think this is more-or-less in René's (deleted) post, but here it is.
First suppose $B\rightarrow C$ is separable. A $\mathbb{F}_q$-point of $B$ or $B'$ must live above a $\mathbb{F}_q$-point $x$ of $C$. If $x$ is a branch point, then both $B$ and $B'$ has a $\mathbb{F}_q$-point above it. Otherwise, the two points of $B$ and $B'$ above $x$ must be defined over $\mathbb{F}_{q^2}$. Suppose a point of $B$ above is defined over $\mathbb{F}_q$, then so is the other one because there is nowhere its Galois conjugate can go. Now for me quadratic twist means to twist the Galois action of $\text{Gal}(\mathbb{F}_{q^2}/\mathbb{F}_q)$ by the only $C$-automorphism of $B$. So the non-trivial element in  $\text{Gal}(\mathbb{F}_{q^2}/\mathbb{F}_q)$ has to send a point on $B'$ above $x$ to the other, i.e. neither of them are defined over $\mathbb{F}_q)$. Similarly when the points in $B$ above $x$ are not defined over $\mathbb{F}_q$, the two in $B'$ must be defined over $\mathbb{F}_q$. So in any case we always have exactly two $\mathbb{F}_q$-points in $B$ and $B'$ above a $\mathbb{F}_q$-point of $C$.
If $p=2$ and $B\rightarrow C$ is inseparable, I guess in this case one has to define $B'=B$. $B$ will be the Frobenius twist of $C$ and thus have the same number of $\mathbb{F}_q$-points as $C$.