Quadratic twist of curve defined over finite field 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 $\#B(\mathbb{F}_q)+\#B'(\mathbb{F}_q)=2\#C(\mathbb{F}_q)$? 
 A: [Edit: there was an obvious mistake in my original answer, which was noticed in the comments. Here's the amended statement.] Let $P$ be a rational point on $C$, then if $P$ is not a branch point then there are two rational points over $P$ on precisely one of $B$ and $B'$ and zero on the other; if $P$ is a branch point, then on both $B$ and $B'$ there is a single rational point over $P$.
A: 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 $C$.
A: Here is an explanation via explicit equations.  First suppose $q$ is odd.  Since the function field extension $\mathbf{F}_q(B)/\mathbf{F}_q(C)$ has degree $2$, it is the extension gotten by adjoining to $\mathbf{F}_q(C)$ the square root of some nonsquare element $\,f\in\mathbf{F}_q(C)$.  Geometrically this means that $\,f\colon C\to\mathbf{P}^1$ is a nonconstant rational function on $C$, and $B$ is the curve defined by (the equations defining the condition that $x$ is a point on $C$ and) $\,y^2=f(x)$, and then the cover $B\to C$ is given by $(x,y)\mapsto x$.  Now the quadratic twist of this cover is the projection $(x,z)\mapsto x$ mapping $B
'\to C$, where $B'$ is defined by the equations $x\in C$ and $z^2=n\cdot f(x)$, with $n$ being any prescribed nonsquare in $\mathbf{F}_q$.  (Different choices of nonsquares $n$ yield isomorphic covers of $C$.)  Finally, pick a point $P\in C(\mathbf{F}_q)$, and let $r$ be the order of vanishing of $\,f$ at $P$.  If $r$ is odd then $P$ has ramification index $2$ in both $B\to C$ and $B'\to C$, so that $P$ lies under a unique point of each of $B(\mathbf{F}_q)$ and $B'(\mathbf{F}_q)$.  Now suppose $r$ is even, and let $t\colon C\to\mathbf{P}^1$ be a function over $\mathbf{F}_q$ which vanishes at $P$ to order $1$.  Let $u$ be the function $\,f/t^r$, so that $u$ has neither a zero nor a pole at $P$, whence $u(P)\in\mathbf{F}_q^{\times}$.  If $u(P)$ is a square in $\mathbf{F}_q^{\times}$ then $P$ has two distinct preimages in $B(\mathbf{F}_q)$ but no preimages in $B'(\mathbf{F}_q)$.  If $u(P)$ is a nonsquare in $\mathbf{F}_q^{\times}$ then $P$ has two distinct preimages in $B'(\mathbf{F}_q)$ but no preimages in $B(\mathbf{F}_q)$.  Therefore in every case, $P$ lies under a combined total of two points from $B(\mathbf{F}_q)$ and $B'(\mathbf{F}_q)$.
You can do the same sort of thing if $q$ is even.  In that case the degree-$2$ cover is defined by $y^2+y=f(x)$, and its quadratic twist is $y^2+y=n+f(x)$, where $n\in\mathbf{F}_q(C)$ cannot be written as $u^2+u$ with $u\in\mathbf{F}_q(C)$.
[Added later: I implicitly assumed that the cover is separable when $q$ is even.  If it is inseparable then $\#B(\mathbf{F}_q)=\#C(\mathbf{F}_q)$.]
