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?
On
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 $uI + vM.$
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.