Show the polynomial $f(x)$ is not primitive.

Question

Let $f(x)\in\mathbb{F}_p[X],\deg f=n>2$, assume that $$f(x)=\sum_{i=0}^n a_i x^i, f(0)\neq 0,$$ and $$x^nf(\frac{1}{x})=f(x).$$ Show that $f(x)$ is not a primitive polynomial.

Let $f(x)\in\mathbb{F}_p[X]$, the minimum positive integer $l$ such that $f(x)| x^l-1$ is called the period of $f(x)$. And such $l$ exists if and only if $f(0)\neq 0$.

For an irreducible polynomial $f(x)\in\mathbb{F}_p[X]$ of degree $n$, we say $f(x)$ is primitive if its period is $l=p^n-1$.

Some thoughts that may be helpful:

1. The last condition is equivalent to say that $a_i=a_{n-i}$.

2. We know that an irreducible polynomial $f(x)\in\mathbb F_p[X]$ of degree $n$ is primitive if and only if the root of $f(x)$ in $\mathbb F_{p^n}$ has order $p^n-1$.

3. If we let $f^*(x):=x^n f(\frac{1}{x})$ for a polynomial $f(x)\in \mathbb{F}_p[X]$ of degree $n$, then we have

• $f(x)$ is irreducible $\Leftrightarrow$ $f^*(x)$ is irreducible.
• If $f(x)$ is irreducible and $f(0)\neq 0$, then $f(x), f^*(x)$ have the same period.

I have no idea how to go on, any help would be appreciated.

Answer 1

The other roots of $f$ are the orbit of $x$ under the Frobenius $$x, x^p, x^{p^2}, ..., x^{p^{n-1}}$$ Since f is palindromic, for each root $\alpha$ of $f$, $1/\alpha$ is also a root, so you have more roots $$x^{p^{n}-1-1}, x^{p^{n}-1-p}, x^{p^{n}-1-p^2}, ..., x^{p^{n}-1-p^{n-1}}$$ and note that $x^i$ for $i\in\{1, ..., p^{n-1}-1\}$ are all distinct. So too many roots.

Answer 2

Hint 0

Assume $f$ irreducible, otherwise it cannot be primitive.

Hint 1

If $\alpha$ is a root of $f$, so is $\alpha^{-1}$.

Hint 2

The relevant Galois group (which is generated by the Frobenius automorphism) acts transitively on the roots of an irreducible polynomial, hence there is $0 \le k < n$ such that $\alpha^{p^{k}} = \alpha^{-1}$.

Hint 3

It follows that $\alpha^{p^{k} + 1} = 1$. Since $\alpha$ has order $p^{n} - 1$, and $k < n$, we must have $p^{k} + 1 = p^{n} - 1$, which holds for $p = 2$, $k = 1$, $n = 2$. The last condition contradicts the assumption $n > 2$.

Hint 4

For $p = 2$, the polynomials $x + 1$ and $x^{2} + x + 1$ are primitive.