Over an algebraically closed field $k$, every one-dimensional torus embedded (as a closed algebraic subgroup) into GL2 is diagonalisable, and the embedding is $t\mapsto (t^m,t^n)$ for some integers $m,n$ with $gcd(m,n)=1$.


What happens when the field $k$ is not algebraically closed? Can I find somewhere the classification of all subtori of GL2, up to conjugation?


By a torus I of course mean a closed algebraic subgroup of GL2, which is isomorphic to $K^{\*}$ over the algebraical closure $K$ of $k$.


One example is, when $k=\mathbb{R}$, the group of rotations. This group is algebraically given by matrices of the form $[a,b,c,d]$ (sorry it seems that matrices do not appear correctly, but you should see what I mean) with $a=d$, $b=-c$ and $a^2+b^2=1$. The same example works over any field, and it is not diagonalisable if $-1$ is not a square in the field.


I would like to have a complete classification.



Classification of Tori of GL2, up to conjugation
In general, a group that is isomorphic to a $d$-dimensional torus over the algebraic closure of the ground field can be identified by the action of the Galois group on its character group / group of one-parameter subgroups, which is $\mathbb Z^d$, so you get a homomorphism $Gal(\bar{k}/k) \to GL\_d(\mathbb Z)$. In the one-dimensional case, this is just a quadratic character, so tori correspond to quadratic fields, in the manner discussed in the comments. In particular, all $1$-dimensional tori in any group embed into $GL\_2$.


The reason you can recover it from the Galois action on the character group is that the points of the torus are exactly the homomorphisms from the character group to $GL\_1(\bar{k})$ that are Galois-equivariant. This is a version of Pontryagin duality.


This is part of the general theory that twists of an object $X$ over a field $k$ are classified by the Galois cohomology group $H^1(k,X)$, or, for twists over an arbitrary base, an etale cohomology group.


EDIT: For a very explicit description, the Galois rep associated to the quadratic field extension $\mathbb Q(\sqrt{D})$ is the group of determinant $1$ matrices that preserve the quadratic form $x^2-Dy^2$, which means


$\left(\begin{array}{cc} a & Db \\ b & a \end{array}\right) $


for $a^2-D b^2 =1 $


for $D=-1$ this reduces to the classic $SO\_2$, and for $D=1$ this is just a regular split torus.


