Given a finite field F, is there a nice way to construct a finite field of pairs of elements of F? For any finite field $F$, there is a finite field created by ordered pairs of elements of $F$. This is because if $F$ has order $p^k$, there are $p^{2k}$ pairs, and so you can make it a field. (This is actually true of fields in general, since for a field of order $k$ for infinite $k$, $k*k=k$.)
Is there a field operation on pairs of $F$ elements in terms of $F$'s operations?
 A: First, pick a quadratic polynomial $f(x)$ over $F$ that has no zero in $F$.  (If the characteristic of $F$ is not 2, then $x^2-a$ will do for suitable $a\in F$, because only half of the nonzero elements of $F$ are squares in $F$.  If the characteristic is 2, then I think something of the form $x^2+x+a$ will do.) Then imagine formally adjoining a root $r$ of $f(x)$ to $F$.  In the resulting quadratic extension of $F$, each element is uniquely of the form $a+br$ with $a,b\in F$.  Identify $a+br$ with the pair $(a,b)$ and you've got a field structure on the set of pairs.
A: same thing, really, find irreducible $x^2 + bx + c.$ Define the matrix
$$
M =
\left(
\begin{array}{cc}
0 & 1 \\
-c & - b
\end{array}
\right).
$$
This is called the companion matrix for the polynomial.
A pair of field elements $(x,y)$ is interpreted as
$$ xI + yM.  $$
Addition is evident, multiplication involves the adjustment coming from $M^2 = -cI - b M.$ The result of multiplication is then another matrix of type $uI + vM.$
