Basis over GF(2^5) with sum of its elements = 1 I need to work over $GF(2^5)$ starting from $GF(2)$ and I would need a basis $\{a_0, a_1, \cdots , a_4\}$ over $GF(2^5)$ such that $a_0+a_1+\cdots +a_4=1$. Do you know how could i proceed? 
Thank you! 

The sums of the roots of the primitive polynomials 
$$x^5+x^4+x^3+x^2+1, \quad x^5+x^4+x^3+x+1, \quad x^5+x^4+x^2+x+1\tag{1}$$
in $\mathbb F_2[x]$ all equal $1$ (the coefficient of $x^4$ is $1$ in each case) and the five roots are a normal basis of $\mathbb F_{32}$ over $\mathbb F_2$.  If you are using $x^5+x^2+1$ for constructing $\mathbb F_{32}$ and denoting the roots as $\alpha, \alpha^2, \alpha^4, \alpha^8, \alpha^{16}$, then these roots sum to $0$ as you have discovered. The roots of the polynomials in $(1)$ above are respectively 
$x^5+x^4+x^3+x^2+1\colon \quad\alpha^3, \alpha^6, \alpha^{12}, \alpha^{24}, \alpha^{17}$
$x^5+x^4+x^3+x+1\colon \quad\alpha^{11}, \alpha^{22}, \alpha^{13}, \alpha^{26}, \alpha^{21}$
$x^5+x^4+x^2+x+1\colon \quad\alpha^5, \alpha^{10}, \alpha^{20}, \alpha^{9}, \alpha^{18}$
and each set is a basis that sums to $1$ as desired. The first set listed above is the one that the OP found for himself, as mentioned in his comment on his own question.
