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Let $q$ be a power of an odd prime. Consider the affine curve $\mathcal C$ defined over $\mathbb F_q$ by $y^2=\prod_{\xi\in\mathbb F_q}(x-\xi)$. I try to determinate the $\mathbb F_q$-automorphism group of $\mathcal C$. I easily found these automorphisms:

$\sigma: (x,y)\mapsto (x+\xi,\pm y)$ with $\xi\in\mathbb F_q$. But I do not know whether there are another ones.

Any idea?

Thanks in advance

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There are other $\mathbf{F}_q$-automorphisms of $\mathcal{C}$. All such are $(x,y)\mapsto (a^2x+b,ay)$ where $b\in\mathbf{F}_q$ and $a\in\mathbf{F}_q^{\times}$.

Here is a description of the automorphism group over the algebraic closure $\Omega$ of $\mathbf{F}_q$ of the projective closure $\mathcal{D}$ of your affine curve $\mathcal{C}$. The projection $\pi\colon\mathcal{D}\to\mathbf{P}^1_x$ has degree $2$, and if $q\ge 5$ then this is the unique degree-$2$ function on $\mathcal{D}$, up to post-composing with automorphisms of $\mathbf{P}^1_x$ (since any curve of genus at least $2$ admits at most one such function). Thus every automorphism $\sigma$ of $\mathcal{D}$ induces an automorphism of $\mathbf{P}^1_x$, so it maps $x\mapsto \mu(x)$ for some degree-one $\mu(X)\in\Omega(X)$. Also $\sigma$ must permute the branch points of $\pi$, namely the points in $\mathbf{P}^1(\mathbf{F}_q)$. For any field $K$ containing $\mathbf{F}_q$ there is a unique degree-one function in $K(X)$ which maps $0$, $1$, and $\infty$ to any prescribed images in $K$, so since $\mu$ permutes $\mathbf{P}^1(\mathbf{F}_q)$ we must have $\mu(X)\in\mathbf{F}_q(X)$. Now $\sigma(y)^2=\sigma(x^q-x)=\mu(x)^q-\mu(x)$ is in $\mathbf{F}_q(x)$, and $\sigma(y)=r(x)y+s(x)$ for some $r,s\in\Omega(X)$, and substituting shows that $s(x)=0$ and $r(x)^2(x^q-x)=\mu(x)^q-\mu(x)$. It follows that the only automorphisms are $$ (x,y)\mapsto \Bigl(\frac{ax+b}{cx+d},\,\frac{y\sqrt{ad-bc}}{(cx+d)^{(q+1)/2}}\Bigr) $$ where $a,b,c,d\in\mathbf{F}_q$ and $ad\ne bc$.

Finally, if $q=3$ then one can verify directly that there are no further $\mathbf{F}_q$-automorphisms of $\mathcal{C}$, but in this case $\mathcal{D}$ has genus $1$ and so has infinitely many automorphisms over $\Omega$.

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    $\begingroup$ Nice! (If anyone else finds this confusing: you get the hyperelliptic involution by choosing $b=c=0$ and $a=d$ a nonsquare in $\mathbf F_q^\times$. Also, if you re-scale the coefficients $a,b,c,d$ by a square in $\mathbf F_q^\times$, then you get the same automorphism.) $\endgroup$ – Dan Petersen May 21 '14 at 8:52

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