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Given that $\mathbb F$ is a field and $\mathbb F[x]$ is the polynomial ring over $\mathbb F$. $\ \ $If the polynomial $a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}$ is irreducible over $\mathbb F[x]$ then so is the polynomial $a_{n}+a_{n-1}x+a_{n-2}x^{2}+\cdots+a_{0}x^{n}$.

Please give me some hints as to how to begin to think the solution.

Thanks for any help.

user26857
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user118494
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    $x^nf(1/x)$ vs. $f(x)$. – Gerry Myerson Aug 28 '15 at 10:11
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    There is the exception that $f(x)=x=1\cdot x+0\cdot1$ is an irreducible polynomial, but $0\cdot x+1\cdot1=1$ is not (because it's a unit of $\Bbb{F}[x]$). Other than this minor exception you can follow the reciprocal route suggested by Gerry, and spelled out in Adayah's answer. – Jyrki Lahtonen Aug 28 '15 at 10:18
  • @GerryMyerson : I still don't get it . $f(1/x)$ does not look like a polynomial though. Some more clue please. – user118494 Aug 31 '15 at 15:26
  • $f(1/x)$ doesn't look like a polynomial, because it isn't a polynomial (if the degree of $f$ exceeds zero). But $x^nf(1/x)$ is a polynomial. – Gerry Myerson Aug 31 '15 at 22:58
  • @GerryMyerson : Yes . But how does that help here? I mean $f(x)$ is irreducible does that imply that $x^{n}f(1/x)$ be so or what ? – user118494 Sep 01 '15 at 06:29
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    If $x^nf(1/x)=g(x)h(x)$ with $g$ of degree $r$, $h$ of degree $s$, $r+s=n$, and then you replace $x$ everywhere with $1/x$, what do you get? – Gerry Myerson Sep 01 '15 at 07:06
  • So, did you try that? – Gerry Myerson Sep 02 '15 at 13:08
  • #GerryMyerson : Yes. I got this . If $x^{n} f(1/x)$=$g(x)f(x)$ then putting ${1}\over {x}$ in place of $x$ everywhere , it is ${{1}\over {x^{n}}}f(x)$= $g(1/x)h(1/x)$ , i.e. $f(x)$=$x^{r} g(1/x).x^{s}h(1/x)$ which gives non-trivial factorization of $f(x)$ , which is a contradiction. So $x^{n} f(1/x)$ cannot have non-trivial factorization. Proved. Right? – user118494 Sep 02 '15 at 15:25
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    Yes, 118494, you got it. Not all that different from the answer @Adayah posted, if you work through the details. I'd encourage you to either post it as an answer (people are encouraged to post answers to their own questions, when the post has led to their understanding how to answer), or to accept Adayah's answer, if you are happy with it. – Gerry Myerson Sep 03 '15 at 00:58

1 Answers1

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Suppose that

$$a_n + a_{n-1} x + \ldots + a_0 x^n = \left( b_i + b_{i-1} x + \ldots + b_0 x^i \right) \left( c_j + c_{j-1} x + \ldots + c_0 x^j \right).$$

Then what is $\left( b_0 + b_1 x + \ldots + b_i x^i \right) \left( c_0 + c_1 x + \ldots + c_j x^j \right)$ ?

Adayah
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