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Let $f$ be a monic primitive polynomial of degree $ab$ over $\mathbb{F}_p$ and let $x$ be a root of $f$.

Let $k = x^{c\frac{p^{ab}-1}{p^b - 1}}$, where $\mathrm{gcd}(c, p^b -1)=1$. Then there is a monic primitive polynomial $h$ of degree $b$ over $\mathbb{F}_p$ such that $h(k) =0$.

What I want is to express $x^a$ in the following way:

$$x^a = \sum_{n=0}^{a-1} \sum_{m=0}^{b-1} {_{n}\textrm{g}_{m}} k^m x^n =: g(k, x) $$

where each $_{n}g_{m} \in \mathbb{F}_p$. If such $g(k,x)$ can be found, then $X^a - g(K, X)$ is a monic primitive polynomial in $X$ of degree $a$ over $\mathbb{F}_{p}[K]/(h) \cong GF(p^b)$.

What I tried initially follows from expressing $g(k,x)$ in the following way:

$$g(k,x) = \sum_{m=0}^{b-1} {g_{m}}(x)k^m$$

where $\mathrm{deg}({g_{m}})< a$. This suggests the following procedure for finding each ${g_{m}}$:

  • Let $d_{b-1}=X^a$
  • Let $K := X^{c\frac{p^{ab}-1}{p^b - 1}}$
  • Starting with $m = b-1$ and decrementing each iteration so that $m = 1$ in the last iteration.
    • Find the inverse of $K^m$ mod $f$, call this polynomial $I_f(K^m)$
    • Let $q^*_{m}$ the product of $I_f(K^m)$ and $d_m$ mod $f$.
    • If $\mathrm{deg}(q^*_{m}) < a$ and $m > 1$, then let $q_m= f + q^*_{m}$, otherwise let $q_{m}=q^*_{m}$
    • Let $g_m = q_m \; \mathrm{mod} \; X^a$ and let $d_{m-1} = q_m - g_m$

The problem with this procedure is that there is no guarantee AFAIK that $f_r := d_{0}/X^a=0$, so then we have:

$$x^a= \sum_{m=0}^{b-1} {g_{m}}(x)k^m + x^af_r(x)$$

We can iterate again with $d_{b-1}=X^af_r$ to replace the $x^a$ term and add all the sums in the end, but again there's no guarantee AFAIK that we will terminate. Is there a way of modifying this procedure to guarantee obtaining $g$ or an entirely different procedure that does the same without trial-and-error?

1 Answers 1

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Using the above notation, this expression allows for the derivation of $g(K,X)$:

$$X^{a} - g(k, X) = \prod_{i\: =\: 0}^{a - 1} (X\; -\; x^{p^{bi}})$$

Here $X$ is the polynomial variable, wile $x$ and $k$ are elements of $GF(p^{ab})$,

Here are some hints to derive this result:

  • If a polynomial over some field has all its roots, then it completely factors into linear terms which correspond to the roots up to multiplicity.
  • For any finite field $GF(p^{n})$, the function $\rho (a) = a^{p}$ is a field automorphism, but not necessarily the identity.
  • Given a finite field $GF(q)$, where $q = p^{b}$, all of its elements are roots of the polynomial $X^{q} - X$.

See Parts I and II of this PDF for more information.