For a finite field GF(2^m), there are 2^m polynomials and the degree of each is no more than m-1.
Can this be applied to GF(5^2)? Would GF(5^2) have 25 elements where each element has a degree no greater than 1?
Would said 25 elements be:
(0, 1, 2, 3, 4)
(D, D+1, D+2, D+3, D+4)
(2D, 2D+1, 2D+2, 2D+3, 2D+4)
(3D, 3D+1, 3D+2, 3D+3, 3D+4)
(4D, 4D+1, 4D+2, 4D+3, 4D+4)
?
I had to look twice, thrice, more, to understand what you were saying. Here’s a restatement, with the field GF(q) denoted instead as $\Bbb F_q$:
If $\Bbb F_{2^m}=\Bbb F_2(\rho)$, i.e. if $\rho$ is an element of $\Bbb F_{2^m}$ whose minimal $\Bbb F_2$-polynomial is of degree $m$, then every element of $\Bbb F_{2^m}$ may be written $a_0+a_1\rho+\cdots+a_{m-1}\rho^{m-2}$, the $a_i$ all lying in $\Bbb F_2$. So the polynomial expressions you were talking about must be polynomials in a special kind of element $\rho$ of $\Bbb F_{2^m}$.
This is basic field theory, and applies, mutatis mutandis to any finite-degree field extension, in particular to $\Bbb F_5\subset\Bbb F_{25}$. So what you say is correct, as long as $D$ is root of an irreducible quadratic polynomial over $\Bbb F_5$. For instance, your polynomial could be $X^2-3$, since $3$ is not a square in $\Bbb F_5$.