Subfields of Matrix Rings The set 
$$S= \left\{ \left(\begin{array}{cc}a&b \\ -b&a \end{array}\right):a, b\in\mathbb{R} \right\}$$
is a subring of the matrix ring $M_2(\mathbb R)$ isomorphic to $\mathbb C$. Can we find other subring $L$ of $M_2(\mathbb R)$ which is also isomorphic to $\mathbb C$ and such that $S\cap L=
\left\{ \left(\begin{array}{cc}a&0 \\ 0&a \end{array}\right):a \in\mathbb R\right\}$?
 A: To describe the possible $L$, you need only say what is the matrix $A$ corresponding to $i$ under the isomorphism with $\mathbb C$. We must have $A^2+I=0$, of course, and in fact this is enough: given any matrix $A$ with that property, the subspace of $M_2(\mathbb R)$ spanned by the identity matrix and $A$ is a field isomorphic to $\mathbb C$ satisfying your condition.
So we need only describe the possible matrices $A$. Can you do that?
A: For any invertible matrix $B$, the conjugation $X \mapsto BXB^{-1}$ is an injective ring homomorphism and therefore maps $S$ to an isomporphic subfield. Try, for example, $B=\begin{pmatrix}1&1\\ 0& 1 \end{pmatrix}$.
A: There is an important link with geometry as follows. The group $\text{GL}_2^+(\Bbb R)$ of matrices with positive determinant acts on the complex upper halfplane
$$
\cal H=\{z=x+iy\in\Bbb C\text{ such that }y>0\}
$$
by linear fractional transformations
$$
\left(\begin{array}{cc}a&b \\ c&d \end{array}\right)\cdot z=\frac{az+b}{cz+d}
$$
Parenthetically, these transformations are isometries when $\cal H$ is given the metric that makes it a model of the hyperbolic plane.
Then the matrices
$$
M=\left(\begin{array}{cc}a&b \\ -b&a \end{array}\right)
$$
are exactly those that stabilize $i$, i.e. $M\cdot i=i$, and form a subgroup isomorphic to the multiplicative group $\Bbb C^\times$. It turns out that the other subgrous of $\text{GL}_2(\Bbb R)$ isomorphic to $\Bbb C^\times$ are precisely the stabilizers of the various points of $\cal H$.
Finally, the fact all these subgroups are conjugated is the "translation" of the fact that the described action is transitive, as
$$
\left(\begin{array}{cc}y&x \\ 0&1 \end{array}\right)\cdot i=x+iy.
$$
