# The largest field contained in $GL_2(\mathbb{R})$

It seems pretty intuitive that the complex numbers can be represented as elements in $GL_2(\mathbb{R})$ by identifying $$a+bi \equiv \begin{bmatrix} a & -b\\b & a \end{bmatrix}$$ And additionally including the matrix $\begin{bmatrix} 0&0\\0&0 \end{bmatrix}$ since it is not invertible.

So it appears that this set in fact forms a field under matrix addition and matrix multiplication.

My question is, can we say that this is the largest field we can obtain out of elements of $GL_2(\mathbb{R}) \cup \{0\}$? If yes, would there be some extension of such a procedure to higher dimensions for say $GL_n(\mathbb{R})$ in which we can get fields out of $\mathbb{R}$-vector spaces? Or can it be shown that it is not possible?

Some initial thoughts on the case for $GL_2(\mathbb{R})$:

I can see that $GL_2(\mathbb{R})$ itself does not form a field since it is not closed under addition and additional constraints would be needed to ensure $det(A + B)$ is non-zero given $det(A)$ and $det(B)$ are non-zero. The condition that:

$$det (A_1 + A_2) = det(\begin{bmatrix} a_1&b_1\\c_1&d_1 \end{bmatrix} + \begin{bmatrix} a_2&b_2\\c_2&d_2 \end{bmatrix}) \neq 0$$

Seems to give me: $$\det(A_1) + det(A_2) \neq -(a_1d_2 - b_1c_2 + a_2d_1 - b_2c_1)$$ Which seems like a pretty odd condition since it would have to be true regardless of choice of $A_1$ and $A_2$. Not sure if this is the right approach...

Maybe we could obtain a lower bound on $det(A_1 + A_2)$ in terms of $det(A_1)$ and $det(A_2)$ by restricting ourselves to matrices of positive determinant along with some extra condition. Clearly skew symmetric matrices with identical diagonal entries works, so maybe positive entries on the diagonal plus skew-symmetric is a step forward?

Actually, we can't obtain a larger field out of elements of $GL_2(\Bbb R)$. Moreover, any field that can be obtained by adding $0$ to a subgroup of $GL_n(\Bbb R)$ is isomorphic to $\Bbb R$ or $\Bbb C$. The reason for this is that $\mathbb{C}$ is an algebraic closure of $\mathbb{R}$; and it can be proven that any algebraic (and in particular finite-dimensional) extension of $\Bbb R$ can be embedded in $\Bbb C$. In particular, such an extension has dimension at most $2$, and thus it is either $\Bbb R$ or $\Bbb C$.