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I'm trying to show that $f(x)=1+x^p+x^{2p}+\dots +x^{p(p-1)}$ is irreducible over $\mathbb{Q}[X]$. I'm well aware that cyclotomic polynomials are irreducible, however the (many) proofs of this statement are quite involved. Since this exercise appears in a textbook that only covers basic algebra, there should be a more direct approach to this exercise.

I tried all criterions that I know to solve this (Eisenstein, linear substitutions, reducing to $\mathbb{Z}_p[X],\dots$). Only one remains. We know that $f(x)=\frac{1-x^{p^2}}{1-x^p}$, or equivalently $1-x^{p^2}=(1-x^p)f(x)$. We know that the roots of $1-x^{p^2}$ are given by $$e^{\frac{2\pi i k}{p^2}}, \mbox{ for $0\leq k\leq p^2-1$ }. $$ Similarly, the roots of $1-x^{p}$ are given by $$e^{\frac{2\pi i kp}{p^2}}, \mbox{ for $0\leq k \leq p-1$}.$$

Hence the roots of $f(x)$ are given by $e^{\frac{2\pi i k}{p^2}}$ where $p\not| \;k$ and $0\leq k\leq p^2-1$. One way to show that $f(x)$ is irreducible over $\mathbb{Q}$ is to show that any product of linear factors of $f(x)$ is not a polynomial over $\mathbb{Q}$. I tried to show this, but it is too difficult.

Does anyone know a elementary way of solving this exercise?

Mathematician 42
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1 Answers1

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We have $f(x)(x^p-1)=x^{p^2}-1$. Reduce mod $p$ and apply $x \mapsto x+1$. We get

$$f(x+1)((x+1)^p-1)=(x+1)^{p^2}-1.$$

Using $(x+1)^p=x^p+1$ and $(x+1)^{p^2}=x^{p^2}+1$ in $\mathbb F_p[x]$, we get

$$f(x+1)x^p=x^{p^2},$$

i.e. $f(x+1) = x^{p(p-1)}$ in $\mathbb F_p[x]$.

The constant term of $f(x+1)=1+(x+1)^p+(x+1)^{2p} + \dotsb + (x+1)^{p(p-1)}$ can be easily seen to be $p$, since there are $p$ summands and each summand contributes a $1$ to the constant term.

Lets summarize: All non-leading terms vanish $\mod p$ and the constant term is not divisible by $p^2$. Hence Eisenstein can be applied to $f(x+1)$ and we are done.

MooS
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