Let $k$ be an algebraically closed field, and let $k^*$ be the one dimesional torus. We want to embed it in $SO(2)$ , the group of matrices $A$ such that $\det A=1$ and $A^tA=Id$.
My first attempt was to define a map $$ x \mapsto \begin{pmatrix} x & \sqrt{1-x^2} \\ -\sqrt{1-x^2} & x \end{pmatrix}$$
But this doesn't work because I can't choose/distinguish the two roots (which exist being K alg. closed).
So we wanted to define the embedding using a matrix of this kind $$ x \mapsto \begin{pmatrix} Ax + Bx^{-1} & Cx + Dx^{-1} \\ -(Cx + Dx^{-1}) & Ax + Bx^{-1} \end{pmatrix}$$
using the fact that identity and inverse are automorphism of $K^*$.
We started doing the computations to find suitable $A,B,C,D$ in the field but our computations does not lead anywhere.
So: is it only a matter of computation? or are we missing some basic fact which could come in handy? thanks in advance
Try changing basis.
In the current basis, you are looking for determinant $1$ matrices which preserver the quadratic form $x^2 + y^2$. But over an alg. closed field (of char. different from $2$ --- the char. $2$ case is a bit degenerate for orthogonal groups, and I won't try to address it) you can change basis so that this quadratic form becomes $x^2 - y^2$, and then change basis again so that it becomes $xy$.
(I am sketching the isomorphism between $SO(2)$ and $SO(1,1)$ over an alg. closed field.)
It shouldn't be so hard to embed $k^{\times}$ into the matrices which preserve the quadratic form $xy$.