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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!


marked as duplicate by Jyrki Lahtonen, Joshua Mundinger, Paul Frost, Feng Shao, nmasanta Aug 22 at 1:56

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  • $\begingroup$ Thanks for the edit! $\endgroup$ – Kai Jun 19 at 2:34
  • $\begingroup$ 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? $\endgroup$ – Dionel Jaime Jun 19 at 3:05
  • $\begingroup$ 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 $\mathbb Z[x]$ $\endgroup$ – Kai Jun 19 at 4:03
  • $\begingroup$ 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 $\endgroup$ – Cardboard Box Jun 19 at 5:16
  • $\begingroup$ This follows by @Dane's comment and en.wikipedia.org/wiki/Eisenstein%27s_criterion applied to a shifted version of the polynomial. $\endgroup$ – pre-kidney Jun 19 at 5:29

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.

  • $\begingroup$ $(x-1)f(x)$ should have degree $p$. Should it be $x^p+x^{p-1}+...+x-p$? $\endgroup$ – Kai Jun 21 at 22:11
  • $\begingroup$ Well done. Thank you! $\endgroup$ – Kai Jun 22 at 2:49
  • $\begingroup$ 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. $\endgroup$ – Sil Jun 23 at 9:19
  • $\begingroup$ 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. $\endgroup$ – Kai Jun 23 at 20:40
  • 1
    $\begingroup$ 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. $\endgroup$ – Cardboard Box Jun 24 at 12:02

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