Find $m$ degree $q^{m-1}$ polynomials which give a bijection $\mathbb{F}_{q^m}\cong \mathbb{F}_{q}^m$ with the map $x\mapsto (p_1(x),\dots,p_m(x))$

Question

Let $q$ some prime power. Now take the field $\mathbb{F}_{q^m}$.

We need to find polynomials $p_1(x),p_2(x),\dots,p_m(x)\in \mathbb{F}_{q^m}[x]$ such that $\deg{p_i(x)}=q^{m-1}$ and they also satisfy this conditions: If we construct the map $\phi:\mathbb{F}_{q^m}\to \left(\mathbb{F}_{q^m}\right)^m$ which sends $x\mapsto (p_1(x),\dots,p_m(x))$ it is also a bijection $\mathbb{F}_{q^m}\cong \mathbb{F}_{q}^m$ as a $\mathbb{F}_q$ vector space.

What i am finally getting that we can represent every element of $\mathbb{F}_{q^m}$ as $a_0+a-1x+\cdots+a_{m-1}x^{m-1}$ Then $$p_i(a_0+a_1x+\cdots+a_{m-1}x^{m-1})=a_{i-1}$$ Then the polynomials will satisfy all these conditions.

For that i thought of using the $q^m\times ({q^{m-1}+1})$ matrix. ${q^{m-1}+1}$ because the polynomial have degree $q^{m-1}4 so that many coefficients. But the vander mond matrix equations more than the number of variables so there is a chance that it has no solution.

I dont know how to proceed

Answer 1

This is one way:

Choose a basis $\alpha_1, \dots,\alpha_m$ of $\mathbb{F}_{q^m}$ seen as a $\mathbb{F}_{q}$-vector space. Now take the polynomials $$p_i(x) = \sum_{k=0}^{m-1} (\alpha_i\, x)^{q^k}.$$ To see that this choice works, check the following:

• If $\beta\in\mathbb{F}_{q^m}$ then $p_i(\beta)^q=p_i(\beta)$, so $p_i(\beta)\in\mathbb{F}_q$
• the map $\mathbb{F}_{q^m}\to(\mathbb{F}_{q})^m$, $\beta\mapsto(p_1(\beta),\dots,p_m(\beta))$ is $\mathbb{F}_{q}$-linear
• if $\beta\neq0$ then $(p_1(\beta),\dots,p_m(\beta))\neq 0$ (otherwise we'd have $r(\gamma):=\sum_{k=0}^{m-1} (\gamma\,\beta)^{q^k}=0$ for every $\gamma\in\mathbb{F}_{q^m}$, giving us a polynomial with too many roots)