Let $f$ be a polynomial with a non-zero constant term and define its reciprocal polynomial $f^*=x^nf(\frac{1}{x})$.
(Here $\deg {f}=n$. I see several other names like "reverse/inverse polynomial", and in the following case "self-inverse".)
We say a polynomial $f$ is primitive over a field $F_p$(p is prime) iff $f$ is irreducible and has its degree $n$ with $k=p^n-1$the lowest one satisfies $f|x^k-1$. Please prove the following:

$f=f^*$and $\deg f>2$ $\Rightarrow$ $f$ is not primitive.

I don't have any specific idea. If some properties or definition of the primitive polynomial are used, you can just state it. And hints will also be apprciated.
Thanks in advance.

Edit: assuming $\deg f>2$
Edit2: the definition of "primitive"

  • $\begingroup$ I think it may relate to the roots: $f(u)=0\iff f^*(u^{-1})=0$ and $f(u)=0$ then $ u,u^p \cdots ,u^{p^{n-1}}$ are the roots of $f$. $\endgroup$ Jun 3, 2019 at 8:14
  • $\begingroup$ With the properties above, we can find a solution considering its roots in $\mathbb{F_{p^n}}$. $\endgroup$ Jun 3, 2019 at 9:05

1 Answer 1


Original Answer: The claim is false.

Take $p=2$, the polynomial $x^2+x+1$ is irreducible over $\mathbb{F}_2$ (hence primitive) and self-reverse.

Addendum: With the constraint $\deg f>2$ the result is still false. The self-reverse polynomial $x^4+x^3+x^2+x+1$ is irreducible degree 4 over $\mathbb{F}_2$ so is primitive.

There seems to be something wrong with your definition of primitive. Over finite fields, an irreducible polynomial having a root in a finite extension implies the polynomial splits completely in that extension, we have $f\mid x^{p^n-1}-1$ automatically if $f\in\mathbb{F}_p[x]$ irreducible degree $n$.

Addendum 2: Now with the new definition of primitive it is easy. Every self-reverse irreducible polynomial $f$ of degree $2n$ over $\mathbb{F}_p$ is actually a factor of $$ x^{p^n+1}-1\in\mathbb{F}_p[x] $$ (in fact also works for $q$) because if $\lambda$ is a root of $f$, then $\lambda^p, \lambda^{p^2}, \dots, \lambda^{p^{2n-1}}$ are the other roots of $f$. Because this set is invariant under inversion, $\lambda^{-1}=\lambda^{p^j}$ for some $j\in\{0,1,\dots,2n-1\}$. Do that again, $$(\lambda^{p^j})^{-1}=(\lambda^{-1})^{p^j}=(\lambda^{p^j})^{p^j}=\lambda^{p^{2j}},$$ so we must have $2j=2n$ and hence $f\mid x^{p^n+1}-1$.

  • $\begingroup$ Sorry, I missed something. I'll edit the question. $\endgroup$ Jun 3, 2019 at 7:59
  • $\begingroup$ Still doesn't work, see edit/addendum $\endgroup$ Jun 3, 2019 at 8:26
  • $\begingroup$ Thanks, there is something wrong with the definition I wrote. I'll edit it. Sorry for my carelessness. $\endgroup$ Jun 3, 2019 at 8:48
  • 1
    $\begingroup$ +1 for a correct and nice argument. FYI this notion of primitive is standard in the context of finite fields. E.g. Lidl & Niederreiter uses it. See also my answer here. $\endgroup$ Jun 3, 2019 at 10:53
  • 1
    $\begingroup$ It may also be worth pointing out that there are no odd degree ($>2$) irreducible palindromic polynomials as such a polynomial has $-1$ as a zero. $\endgroup$ Jun 3, 2019 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .