Find normal basis for F390625 over F25.

Question

I am new to Field theory and want to do this problem , Can anyone give hint or reference on how to find Normal basis, I tried searching over the web but did not find any good resources

Answer 1

Let $K=\Bbb F_{5^2}$ and $L=\Bbb F_{5^8}$. Then $|L:K|=4$ and $\text{Gal}(L/K)$ is cyclic, generated by the Frobenius map $F:a\mapsto a^{25}$.

We get an action of the polynomial ring $K[X]$ on $L$ by defining $$\left(\sum_i c_iX^i\right)\cdot a=\sum_i c_iF^i(a).$$ Then $L$ is annihilated by $X^4-1$ in this action

For $a\in L$ to be a normal basis, it is necessary and sufficient that the annihilator of $a$ in $K[X]$ is generated by $X^4-1$. The annihilator is a principal ideal, and is generated by a monic factor of $X^4-1$. But $X^4-1$ factors completely over $K$: $$X^4-1=(X-1)(X+1)(X-i)(X+i)$$ where $i^2=-1$ in $K$. So $a$ is a normal basis iff $$\frac{X^4-1}{X-u}\cdot a\ne0$$ for each $u\in\{\pm1,\pm i\}$.

One can solve this by trial and error: keep throwing random $a$ at this until one of them works. But we can use some more algebra to avoid work, as $K$ has the fourth roots of unity, $L/K$ is a Kummer extension: $L=K(\sqrt[4]c)$ where $c$ is any non-square in $K$. The typical element of $L$ is $$a=b_0+b_1\sqrt[4]c+b_2(\sqrt[4]c)^2+b_3(\sqrt[4]c)^3$$ with $b_0,\ldots,b_3\in K$. Then $$F(a)=b_0+jb_1\sqrt[4]c-b_2(\sqrt[4]c)^2-jb_3(\sqrt[4]c)^3.$$ where $j$ is either $i$ or $-i$. Repeating this, we see that $a,F(a),F^2(a),F^3(a)$ are linearly independent iff $b_0b_1b_2b_3\ne0$.

Answer 2

Apply the iterative construction of a normal basis for the field extension $\Bbb F_{p^n} \mid \Bbb F_p$ for $p=5$, described on page $42$.