# Proof of irreducibilty of $f(x)=\sum_{i=0}^{p-1}(p-i)x^i$ [duplicate]

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, nmasantaAug 22 at 1:56

• Thanks for the edit! – Kai Jun 19 at 2:34
• 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 at 3:05
• 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 at 4:03
• 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 at 5:16
• This follows by @Dane's comment and en.wikipedia.org/wiki/Eisenstein%27s_criterion applied to a shifted version of the polynomial. – 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.
• $(x-1)f(x)$ should have degree $p$. Should it be $x^p+x^{p-1}+...+x-p$? – Kai Jun 21 at 22:11
• 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 at 9:19
• 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 at 20:40
• 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 at 12:02