Characterize all sets of $2\times 2$ real matrices isomorphic to $\mathbb C$ as a field

Question

Let $A$ be the set of all 2x2 real matrices of the form $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$ Show that a set $B$ of $2\times 2$ real matrices is isomorphic to $\mathbb C$ as a field iff $B=gAg^{-1}$ for some invertible real matrix $g$.

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I've shown that any field of the above form is isomorphic to $\mathbb C$, but I cannot figure out how to prove the other direction, that if $B$ is isomorphic to $\mathbb C$ then it is of the above form. I can't find any way to characterize what $g$ would have to be for a given $B$. The statement is basically equivalent to saying the set $B$ is the set of all matrices representing the linear operators on the vector space $\mathbb C$ over $\mathbb R$ of the form $T_w(z)=wz$ under some basis, but I cannot prove that either.

Help would be appreciated.

Answer 1

Two caveats, and some high-level hints:

First, the claim you're asked to prove is not necessarily true! There exist proper subsets of $\mathbb C$ that are field-isomorphic to $\mathbb C$ itself, and such a set will map to a proper subset of $A$ that is isomorphic to $\mathbb C$ but certainly doesn't have the shape specified for $B$ in the problem.

(To prove that such a subset exists requires the axiom of choice, but it won't help us to decide to work in ZF -- if we prove the claim from ZF we would have disproved the axiom of choice, and a result of Gödel (no, not the famous one) states that the axiom of choice cannot be disproved in ZF -- unless ZF is inconsistent, in which case they sky is falling anyway).

In order to fix this and get something that can be proved, we need to revise the claim in one of two ways:

1. Replace "isomorphic to $\mathbb C$ as a field" with "isomorphic to $\mathbb C$ as an $\mathbb R$-algebra" -- that is, require that scaling a matrix in $B$ by a real constant corresponds, via the isomorphism, to multiplying the complex number by that same real number. Or,

2. Require that $B$ is a closed subset of $\mathbb R^{2\times 2}$.

Second, it is not really clear just what was meant by "isomorphic to $\mathbb C$ as a field" in the first place. For example, if $D$ is the set of $\mathbb 3\times 3$ matrices of the form $$ \begin{bmatrix}a & -b & 0 \\ b & a & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ should we then consider $D$ to be "isomorphic to $\mathbb C$ as a field"? It definitely is isomorphic to $\mathbb C$ as an $\mathbb R$-algebra, but $1\in\mathbb C$ is not represented by the identity element of $\mathbb R^{3\times 3}$. Now, the concept of a field homomorphism does require that $1$ maps to $1$, but $\mathbb R^{3\times 3}$ is not a field, and it is not a priori clear whether or not viewing a subset of it as a field should require us to use the particular identity element from the surrounding non-field.

Back in the $2\times 2$ case: We assume that an isomorphism $\varphi:\mathbb C\to B$ has been given. If we can't assume that $\varphi(1) = I_{2\times 2}$, then the solution will need some additional footwork to get started. In particular, then we don't automatically know that $\varphi(i)$ must be a solution of $X^2 = -I_{2\times 2}$. We can in fact still get through by instead starting with $X^3 = -X$ (together with $X\ne 0$), but there are additional details to keep track of in that case.

In each of these cases, some thinking about either $X^2 = -I$ or $X^3 = -X$ should enable you to find an invertible basis change $g$ such that $$X = g\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}g^{-1}.$$ If you don't yet know that $\varphi(1) = I_{2\times 2}$, computing $-X^2$ in terms of $g$ will show it now.

If you have decided that $B$ must be an $\mathbb R$-algebra and isomorphic to $\mathbb C$ as such, it is now easy to conclude that $B$ is in fact identical to $gAg^{-1}$.

However, if all you know is that $B$ is closed, it's not quite that straightforward. Now that you know that $I\in B$, you can show that $\lambda I \in B$ for every real $\lambda$ -- but beware that this doesn't guarantee that $\lambda I=\varphi(\lambda)$! Since $B$ is closed under matrix multiplication and addition, we can then at least conclude $gAg^{-1} \subseteq B$. But an additional argument will be needed to show that $B$ cannot be larger than $gAg^{-1}$. One strategy could be to show that if there is a $b\in B$ with $g^{-1}bg\notin A$, then you can find an nonzero non-invertible element in $B$ -- which certainly cannot represent anything from $\mathbb C$.

Answer 2

As discussed in the other answer, I agree that the way it's phrased right now is not quite correct. I think we should rephrase such that $B$ is a unital $\mathbb R$-subalgebra of $M_2(\mathbb R)$ which is isomorphic as an $\mathbb R$ algebra to $\mathbb C$. Of course $B$, being linearly isomorphic to $\mathbb C$, is a $2$-dimensional $\mathbb R$ vector space.

With that in mind, suppose $B$ is ($\mathbb R$-linearly) ring isomorphic to $\mathbb C$ and define $\tilde I=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $I$ be the identity matrix. Of course, $A$ and $B$ share the same identity $I$, which the isomorphism preserves.

Then $B$ contains an element $J$ whose characteristic (and minimal) polynomial is $X^2+1=0$. It is a well-known exercise that matrices with side length $3$ or less over a field are similar if they have the same characteristic polynomial and minimal polynomial. Then since $J$ and $\tilde I$ have the same characteristic (and minimal) polynomial, there exists an invertible matrix $g$ such that $J=g\tilde Ig^{-1}$.

Since conjugation by $g$ is an injective $\mathbb R$-linear map, it's guaranteed to be onto $B$ (considering dimensions.) It is also a ring homomorphism, so indeed it is an $\mathbb R$-algebra isomorphism of $A$ and $B$.