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Let $p$ be an odd prime. Show that $f(x)=\displaystyle\sum_{i=0}^{p-1}(p-i)x^i$ is irreducible.

Clearly modulo reduction doesn't work (Since this is already modulo $p$ reduced). So I've thought about the roots of this polynomial. Note $f(0)=p, f(-1)=\frac{1+p}{2}$, but then I'm stuck. Any hint would be appreciated!

StubbornAtom
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Kai
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  • Thanks for the edit! – Kai Jun 19 '19 at 02:34
  • If this is already modulo $p$ reduced, (i.e $p =0$) then $x$ is a factor of this polynomial. Should the index be starting at $1$ instead of $0$? Maybe you should clarify what field you want to say it's irreducible over? – Dionel Jaime Jun 19 '19 at 03:05
  • Sorry for the possible confusion. The coefficient of $x_{p-1}$ is 1. Expanded, my polynomial is $x^{p-1}+2x^{p-2}+...+p$. I'm trying to show it is irreducible over $\mathbb Z[x]$ – Kai Jun 19 '19 at 04:03
  • This may not end up helping, but $(x-1)f(x) = x \Phi_p(x) - p$, where $\Phi_p(x) = 1 + x + \cdots + x^{p-1}$ is the $p$th cyclotomic polynomial – Cardboard Box Jun 19 '19 at 05:16
  • This follows by @Dane's comment and https://en.wikipedia.org/wiki/Eisenstein%27s_criterion applied to a shifted version of the polynomial. – pre-kidney Jun 19 '19 at 05:29
  • @pre-kidney That was my initial thought, but it is unclear to me which shift would work. – Cardboard Box Jun 19 '19 at 15:48
  • Maybe try proving $f(x+1)$ is irreducible? – Kai Jun 20 '19 at 18:52
  • Similar to this https://mathoverflow.net/questions/18094/polynomial-with-the-primes-as-coefficients-irreducible , can be solved by the same method. The answer here follows the same path. – Sil Jun 22 '19 at 18:58

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First note that $$ (x-1) f(x) = x^p+x^{p-1} + x^{p-2} + \cdots + x - p . $$ This implies that all the roots of $f$ lie strictly outside the unit circle. For if $f(\alpha) = 0$ with $|\alpha| \leq 1$, then rearranging the above and using the triangle inequality yields $$ p = |\alpha + \cdots + \alpha^{p}| \leq \sum_{i=1}^{p} |\alpha|^i \leq p . $$ But this implies that $\alpha = 1$, which is not the case since $f(1) \neq 0$.

Now suppose that $f$ factors as $f = gh$. Then since $p = f(0) = g(0) h(0)$ is prime, one of $g$ or $h$ must have constant term equal to $\pm 1$. But then $\pm 1$ is the product of the roots of this factor, which are all greater than 1 in absolute value. But this is impossible, so we conclude that $f$ is irreducible.

Kai
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Cardboard Box
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  • $(x-1)f(x)$ should have degree $p$. Should it be $x^p+x^{p-1}+...+x-p$? – Kai Jun 21 '19 at 22:11
  • Well done. Thank you! – Kai Jun 22 '19 at 02:49
  • Any $\alpha$ on the unit circle will satisfy $\sum_{i=1}^{p} |\alpha|^i = p$, I think an additional idea is required to justify $\alpha=1$ particularly. – Sil Jun 23 '19 at 09:19
  • No. For $\alpha=1$, the new function $x^p+x^{p-1}+\cdots+x-p=(x-1)f(x)$, so a factor of $x-1$ doesn't count. Plugging $\alpha=1$ into the original function $f(x)$, you clearly have a positive integer. – Kai Jun 23 '19 at 20:40
  • You didn't get my point, it is about $|\alpha|=1$, the proof at some point concludes that it implies $\alpha=1$ (which is immedietally dismissed by contraditiction). All I am saying is that it needs more work to exlude the roots elsewhere on the unit circle. – Sil Jun 24 '19 at 06:40
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    The conclusion $\alpha = 1$ follows from having equality in both of the inequalities together. We have equality in the "triangle inequality" step $|\alpha + \cdots + \alpha^p| \leq \sum |\alpha|^i$ only if all the powers of $\alpha$ are pointing in the same direction. – Cardboard Box Jun 24 '19 at 12:02
  • Right, that is what was missing. – Sil Jun 24 '19 at 18:18