# Prove reciprocal polynomial is not primitive(field).

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.

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

• 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 $f$. Commented Jun 3, 2019 at 8:14
• With the properties above, we can find a solution considering its roots in $\mathbb{F_{p^n}}$. Commented Jun 3, 2019 at 9:05

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 $$f\mid x^{p^n+1}-1$$.

• Sorry, I missed something. I'll edit the question. Commented Jun 3, 2019 at 7:59
• Still doesn't work, see edit/addendum Commented Jun 3, 2019 at 8:26
• Thanks, there is something wrong with the definition I wrote. I'll edit it. Sorry for my carelessness. Commented Jun 3, 2019 at 8:48
• +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. Commented Jun 3, 2019 at 10:53
• 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. Commented Jun 3, 2019 at 11:02