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?



Seeking conceptual explanation of these nice bijections on roots of unity
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.


