Prime field for linear transformation matrices From Humphreys' Introduction to Lie Algebras and Representation Theory:

Let $\text{End }V$ denote the set of linear transformations $V\rightarrow V$, and we write $\mathfrak{gl}(V)$ for $\text{End }V$ viewed as Lie algebra. We write down the multiplicatio ntable for $\mathfrak{gl}(n,F)$ relative to the standard basis consisting of the matrices $e_{ij}$ (having $1$ in the $(i,j)$ position and  $0$ elsewhere). Since $[e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{li}e_{kj}$, it follows that $[e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{li}e_{kj}$. Notice that the coefficients are all $\pm 1$ or $0$; in particular, all of them lie in the prime field of $F$.

What does it mean to lie in the prime field of $F$? A prime field $GF(p)$ is the field of residue classes modulo $p$. How does it have anything to do with matrices with coefficients $-1,0,1$?
 A: The prime subfield of a field $F$ is the subfield of $F$ generated by the identity element only. It is isomorphic to either $\mathbb{Q}$ (characteristic zero) or $\mathbb{Z}/(p)$ for a prime number $p$ (characteristic $p$).
Reference: http://mathworld.wolfram.com/PrimeSubfield.html
A: The prime field of $F$ is a subfield of $F$, hence contains $0$ and $1$ and also the additive inverse $-1$. 
A: You have a collection of matrices on the one hand, and a collection of modulo classes on the other. The point is that they will be isomorphic to each other.
Consider the field of complex numbers $\mathbb{C}$. There is a subfield of the $\text{Mat}(2,\mathbb{R})$, say $S$, that is isomorphic to $\mathbb{C}$. Consider the following map:
$$\varphi : x + \operatorname{i}\!y \longmapsto \left[\begin{array}{cc} x & -y \\ y & x \end{array}\right]$$ 
You can easily show that $\varphi(z_1+z_2) = \varphi(z_1) + \varphi(z_2)$ and 
$\varphi(z_1z_2) = \varphi(z_1)\varphi(z_2)$. Moreover, $\varphi(z)$ is the identity matrix if and only if $x=1$ and $y=0$, i.e. $z=1$. Hence $\mathbb{C} \cong S$.
On the one hand we have complex numbers, on the other we have some skew-symmetric matrices. It's like what you have. On one hand you have some residue classes and on the other you have some matrices. You'll find that these two sets behave in the same way under your operations.
Note, complex conjugation corresponds to the matrix transpose: $\varphi(\overline{z}) = \varphi(z)^{\top}$.
