Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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?

share|cite|improve this answer
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, $B=\begin{pmatrix}1&1\\ 0& 1 \end{pmatrix}$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.