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 $S\cap L= \left\{ \left(\begin{array}{cc}a&0 \\ 0&a \end{array}\right):a \in\mathbb R\right\}$?

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As a hint, this is very easy to do. Any automorphism of M2R will give you a different C. –  Jack Schmidt Oct 9 '12 at 21:33
@zacarias: Sorry for sticking to details, but saying that some object is a subfield, implicitly assumes that the bigger object is a field. Probably you should say "subgroup which is a field"... –  Dennis Gulko Oct 9 '12 at 21:34
@DennisGulko, the term subfield does not carry the implication that the big thing is a field, in my experience —and I see the term quite often! –  Mariano Suárez-Alvarez Oct 9 '12 at 21:44
@Mariano Suárez-Alvarez: I see it quite often too. I guess it depends on the general context :) –  Dennis Gulko Oct 9 '12 at 21:47
For example, there are whole volumes written about subfields of central simple algebras, and I doubt anyone has ever written «subalgebras which are fields» to refer to them :-) –  Mariano Suárez-Alvarez Oct 9 '12 at 21:49

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?

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There are twice as many such matrices as there are subfields,though: $A$ and $-A$ give the same field, but that is the only issue. –  Mariano Suárez-Alvarez Oct 9 '12 at 21:43

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, $B=\begin{pmatrix}1&1\\ 0& 1 \end{pmatrix}$.

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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 $$\left(\begin{array}{cc}y&x \\ 0&1 \end{array}\right)\cdot i=x+iy.$$

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