Given that $f(x)=\sum\limits_{k=0}^n a_kx^k$ is an irreducible polynomial, show that $g(x)=\sum\limits_{k=0}^n a_kx^{n-k}$ is irreducible too. I don't see how this specific coefficient exchange would preserve the irreducibility. Is there something else I should be looking at?
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2Hint: $g(x)={x^n}f(\frac 1x)$ Now suppose $g(x)$ can be written as $p(x)q(x)$. – lulu Jan 08 '17 at 21:38
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1Of course you want to assume $a_n \ne 0$, otherwise $f(x) = 1 + x + 0 x^2$ is a counterexample. – Robert Israel Jan 08 '17 at 21:47
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1In addition to the objection raised by @RobertIsrael we have the edge case of the irreducible polynomial $f(x)=x$ when the reciprocal $g(x)$ is a unit of the polynomial ring rather than an irreducible element. – Jyrki Lahtonen Jan 09 '17 at 06:53
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Since $f$ is irreducible, $f\neq a_n X^n$, so $\deg g\geq 1$.
Note that $g(X)=X^nf(\frac 1X)$. Supposing $g(X)=r(X)s(X)$ with $\deg r\geq1$, you get $$X^nr(\frac 1X)s(\frac 1X)=f(X)$$ that is $$ X^{n-\deg g} \left(X^{\deg r}r(\frac 1X)\right) \left(X^{\deg s}s(\frac 1X)\right)=f(X)$$
If $\deg (X^{\deg r}r(\frac 1X))\geq 1$ or $\deg (X^{\deg s}s(\frac 1X))\geq 1$ we get a contradiction with the irreducibility of $f$.
Otherwise, $X^{\deg r}r(\frac 1X)$ and $X^{\deg s}s(\frac 1X)$ are both constant, which implies $n-\deg g = \deg f = n$ hence $\deg g = 0$, which is in contradiction with the first line of my answer.
Gabriel Romon
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Hint The reversal map $\, f\mapsto x^d f(x^{-1}),\ d=\deg f\,$ is multiplicative, being the product of two multiplicative maps, namely $\,f(x)\mapsto f(x^{-1}),\,$ and $\,f(x)\mapsto x^{\,\deg f}\ $ (an "exponental" of the additive degree map). Being multiplicative, it preserves (ir)reducibility.
Bill Dubuque
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