# 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 $\#B(\mathbb{F}_q)+\#B'(\mathbb{F}_q)=2\#C(\mathbb{F}_q)$?

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You can find an "elaborate exposition" in Section I.5.3 of Serre's book "Galois cohomology". That displays the generality of this concept, but might not be the easiest introduction to it. Twisting in the case of curves is explained in section X.2 of Silverman's book "The Arithmetic of Elliptic Curves". – Michael Zieve May 16 '14 at 0:53
An elementary exposition of twists in the case of curves over finite fields is Proposition 5.2.8 of Stichtenoth's book "Algebraic Function Fields and Codes". The proof there is essentially Chebotarev's field-crossing argument, and makes the proof of the function field analogue of Chebotarev's density theorem seem quite natural, which might be helpful for thinking about the proof in the number field case. – Michael Zieve Jun 18 '14 at 7:20

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 $\#B(\mathbf{F}_q)=\#C(\mathbf{F}_q)$.]

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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 $C$.

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[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 $P$.

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Maybe I am being stupid - I thought if there is a rational branch point, then both $B$ and $B'$ has one point above it? – Cheng-Chiang Tsai May 16 '14 at 0:21
@Cheng-Chiang Tsai: you are exactly correct. The word "only" makes René's answer wrong, though his answer is the correct explanation over rational points which are not branch points. – Michael Zieve May 16 '14 at 0:24
Oh, you're absolutely right. Thanks! – René May 16 '14 at 0:30