TITLE: For which fields is $\left\{\begin{pmatrix} a & -b \\ b & a \end{pmatrix} \,\middle |\, a, b \in F\right\} \subset M_2(F)$ a field?
QUESTION [5 upvotes]: Consider $R$ the set of matrices of the form 
$$A = \left(\begin{array}{rr}a& -b \\b& a\end{array}\right),$$ where $a$ and $b$ live in a given field $F$.
I know $R$ is a commutative ring with the identity matrix and now want to determine for which of $F = \mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{F}_5, \mathbb{F}_7$ is $R$ a field? Also how to generally characterize the types of fields including which prime fields $\mathbb{F}_p$ that will make $R$ a field?

REPLY [3 votes]: The matrix $\begin{pmatrix} a & -b \\  b & a \end{pmatrix}$ is invertible iff its determinant, $\delta = a^2 + b^2$  is nonzero. In this case, its inverse is $\delta^{-1} \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$,  which still lives in your set.
So, your question is equivalent to: for which fields does the quadratic form $\delta(x,y) = x^2 + y^2$ admit only one zero, namely $(0,0)$? In fancier terms, for which field is the form $x^2 + y^2$ anisotropic?
The answer is that it is for $\mathbb Q$, $\mathbb R$ (obvious, for sign reasons) and $\mathbb F_7$ (the only squares modulo $7$ are $0$, $1$, $2$ and $-3$, so two squares cannot cancel out) and it isn't for $\mathbb C$ ($1^2 + i^2 = 0$) or $\mathbb F_5$ ($1^2 + 2^2 = 5 = 0$).
In general, for a finite field with an odd number $q$ of elements, the answer is that the form is anisotropic if and only if $-1$ isn't a square, iff $q \equiv 3 \,(\mathrm{mod}\,4)$. It's not very hard and very classical, you can probably find this result in your favourite algebra book.
In general, questions about forms being isotropic or anisotropic is one of the main questions in the algebraic theory of quadratic forms, a fascinating subject in itself. If you have access to it, the beginning of Lam's Quadratic Forms over Fields is a marvelous read (and you will se that questions like "is this matrix ring a field?" pop up quite naturally in this theory).