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I proved the following facts by unenlightening calculations. Since the statements are quite clean, I think there should be a conceptual explanation for them, which my proof certainly is not.

Let $q$ be a prime power, and let $\mu_{q+1}$ be the set of $(q+1)$-th roots of unity in the finite field $\mathbf{F}_{q^2}$. If $b\in\mu_{q+1}$ and $c\in\mathbf{F}_{q^2}\setminus\mathbf{F}_q$ then $$ x\mapsto \frac{cx-bc^q}{x-b} $$ maps $\mu_{q+1}$ to $\mathbf{F}_q\cup\{\infty\}$. If $b\in\mu_{q+1}$ and $d\in\mathbf{F}_{q^2}\setminus\mu_{q+1}$ then $$ x\mapsto \frac{x-bd^q}{dx-b} $$ maps $\mu_{q+1}$ to itself. (It is also true that these are the only degree-one rational functions which map $\mu_{q+1}$ to either $\mathbf{F}_q\cup\{\infty\}$ or $\mu_{q+1}$, but I'm mainly interested in understanding the existence.)

I tagged this "group theory" because the first fact vaguely feels like a connection between orbits of a nonsplit torus and a split torus in $\textrm{PGL}_2(q)$. It's tempting to identify $\mathbf{F}_{q^2}$ with $\mathbf{F}_q\times\mathbf{F}_q$, and consider the resulting action of $\textrm{GL}_2(q)$ on $\mathbf{F}_{q^2}$, but I don't see how to go further in this way.

Any suggestions?

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    $\begingroup$ The parameter $b$ doesn't seem to be doing much: we can translate the group $\mu_{q+1}$ by any single element and in the fractions we can divide throughout on top and bottom. So do we lose anything by only considering $b=1$? $\endgroup$ – Marguax Oct 8 '13 at 3:40
  • $\begingroup$ @Marguax: good point! So it suffices to prove the result for $b=1$. Thanks! $\endgroup$ – Michael Zieve Oct 8 '13 at 3:46
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    $\begingroup$ In the 2nd case, using multiplicative translation by $-x \in \mu_{q+1}$ on the target, it is the same to ask for a "conceptual" proof that $f:x \mapsto ((d^q/x)-1)/(dx-1)$ is a bijection of $\mu_{q+1}$ onto itself. Clearly $f(x)=t(x)^{q-1}$ where $t(x)=dx-1$ (since $x^q=1/x$), so this map lands inside $\mu_{q+1}$. And using that $d^{q+1}\ne 1$ we see by cross-multiplying in an equality of fractions that $f$ is injective on $\mu_{q+1}$, hence bijective. Does this just recreate your argument for the 2nd case? $\endgroup$ – Marguax Oct 8 '13 at 4:05
  • $\begingroup$ Likewise, for the 1st one, was your argument that raising to the $q$th-power gives the same fraction (using that $x^q = 1/x$) and that injectivity is a direct inspection? $\endgroup$ – Marguax Oct 8 '13 at 4:10
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    $\begingroup$ Yeah. A lot like the analogue with complex plane: $\mu_{q+1}$ is the unit circle, $\bf{F}_q$ is the real axis, $x\mapsto x^q$ is the complex conjugation that amounts to taking the reciprocal on the unit circle. I'm not sure that I've seen this before. But I have used (and seen it used by other coding theorists in need of evaluating some character sums) the mapping $x\mapsto (c+x)/(c^q+x)$ to parametrize $\mu_{q+1}\setminus\{1\}$ by elements $x\in\bf{F}_q$. Here $c\in\bf{F}_{q^2}\setminus\bf{F}_q$ is a fixed constant. $\endgroup$ – Jyrki Lahtonen Dec 7 '13 at 20:19
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Let $E$ be a curve defined by a singular Weierstrass equation over $\mathbb{F}_q$, where the singularity is a node, say at the origin. Then, Silverman says (Arithmetic of Elliptic Curves, page 46) that $E$ may be written as $$ E: y^2 + A_1 xy - A_2 x^2 - x^3 = 0, $$ where $A_1^2 + 4 A_2$ is not zero. If the two tangent lines at the node are given by $y = a_i x$ for $i=1,2$, then by comparing coefficients we see that the slopes are roots of $t^2 + A_1 t - A_2$.

Suppose this quadratic is irreducible over $\mathbb{F}_q$. Then it splits in $L = \mathbb{F}_{q^2}$. According to Silverman's Prop. 2.5 on page 56 (properly adjusted to not assume algebraic closure, see Exercise 3.5 on page 105), the map $(x,y) \rightarrow (y - a_1 x)/(y - a_2 x)$ gives us an isomorphism of algebraic groups $E_{ns}(L)\cong L^\times$. Part (ii) of the exercise referenced above shows that under this isomorphism, $E_{ns}(K)$ corresponds precisely to those elements of $L^\times$ whose norm to $K$ is $1$. Now $N(x) = x\cdot x^q = x^{q+1}$, so elements of norm $1$ are precisely the $(q+1)th$ roots of unity in $\mathbb{F}_{q^2}$. Since $E$ is singular with a point defined over $\mathbb{F}_q$, its non-singular part is isomorphic to $\mathbb{P}^1(\mathbb{F}_q)$.

So, we have a bijection between the $(q+1)th$ roots of unity in $\mathbb{F}_{q^2}$ and $\mathbb{P}^1(\mathbb{F}_q)$ given by a linear fractional transformation.

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  • $\begingroup$ Thanks, that's exactly the sort of explanation I was looking for. Incidentally, in order that the map $(x:y:z)\mapsto(x:y)$ should give a bijection $E_{ns}(\mathbb{F}_q)\to\mathbb{P}^1(\mathbb{F}_q)$, we need the $a_i$ to be outside $\mathbb{F}_q$, or equivalently we need $A_1^2+4A_2$ to be a nonsquare in $\mathbb{F}_q$. I think this condition is needed in order for your construction to give the desired bijection $\mu_{q+1}\to\mathbb{P}^1(\mathbb{F}_q)$. Also, not surprisingly my bijections of $\mu_{q+1}$ arise by composing one of your bijections with the inverse of another. $\endgroup$ – Michael Zieve Dec 26 '14 at 16:51
  • $\begingroup$ I agree, this assumption comes into my argument when I assume the quadratic $t^2 + A_1t - A_2$ is irreducible over $F_q$. $\endgroup$ – tghyde Dec 26 '14 at 18:14
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The question and the analog to the Cayley map in complex numbers pointed out in the comment by Jyrki Lahtonen is not only an analog, but in fact both are special cases of this more general observation: Let $z\mapsto\bar z$ be an involutory automorphism of a field $F$, and let $E$ be the subfield fixed by $\bar{\phantom{a}}$. Furthermore, set $S=\{z\in F\;|\;z\bar z=1\}$. Then, for $b\in S$, $c\in F\setminus E$, and $d\in F\setminus S$, \begin{equation*} z\mapsto \frac{cz-b\bar c}{z-b} \end{equation*} maps $S$ to $E\cup\{\infty\}$ and \begin{equation*} z\mapsto \frac{z-b\bar d}{dz-b} \end{equation*} maps $S$ to $S$.

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