Irreducibility of $x^{p(p-1)}+x^{p(p-2)}+\cdots+x^p+1$ over $\mathbb Q$

Question

It is well known that the $p$th cyclotomic polynomial $$\Phi_p(x)\ =\ x^{p-1}+\cdots+x+1\ =\ \frac{x^p-1}{x-1}$$

is irreducible over $\mathbb Q$ for prime $p$. The standard trick is to make the transformation $x\mapsto x+1$ and turn the polynomial to one of the form $x^{p-1}+p\cdot q(x)$, where $q(x)$ is polynomial of degree $p-2$ and constant term 1, and apply Eisenstein's criterion.

On p.35 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition) the authors state that the following polynomial is also irreducible over $\mathbb Q$: $$\Phi_p(x^p)\ =\ x^{p(p-1)}+\cdots+x^p+1\ =\ \frac{x^{p^2}-1}{x^p-1}$$

(where $p$, as before, is prime). According to the book the result can also be proved by applying the transformation $x\mapsto x+1$. I tried it but I'm afraid it got me nowhere; neither have I been able to prove it by any other method.

I would be grateful if someone could show me how it is done. Thanks.

Answer 1

In any commutative ring of characteristic $p$ we have $(a+b)^p=a^p+b^p$. Therefore, when reduced modulo $p$, $$ \Phi_p((x+1)^p)=\frac{(x+1)^{p^2}-1}{(x+1)^p-1}\equiv\frac{x^{p^2}}{x^p} =x^{p^2-p}. $$ So all but the leading coefficient of $\Phi_p((x+1)^p)$ are divisible by $p$.

On the other hand $\Phi_p(1^p)=p$, so the constant term of $\Phi_p((x+1)^p)$ is not divisible by $p^2$. Eisenstein kicks in, and the irreducibility of $\Phi_p(x^p)$ follows.

Answer 2

Well, $ \varphi(p^2) = p(p-1) $, so the $p^2 $-th cyclotomic polynomial is of degree $ p(p-1) $, monic, and it divides $ p(x) = (x^{p^2} - 1)/(x^p - 1) $, since any primitive $ p^2 $-th root of unity remains a root of this polynomial. This means that $ p(x) $ is the cyclotomic polynomial, and it is thus irreducible.

The idea of the transformation $ x \to x+1 $ is that the prime $ p $ totally ramifies in $ \mathbf Q(\zeta_{p^2}) $ as $ (p) = (1 - \zeta_{p^2})^{p(p-1)} $. It is then well-known that the characteristic polynomial of $ 1 - \zeta_{p^2} $ must be Eisenstein at $ p $, since it generates the prime ideal lying above $ p $.

Answer 3

Here’s another way of doing the same thing, very similar to what Jyrki has written:

It’s a little clearer, I think, to show that for $n\ge1$, $\Phi_{p^{n+1}}(X)=(X^{p^{n+1}}-1)\big/(X^{p^n}-1)$ is irreducible. We set $$g(X)=\Phi_p(1+X)=\frac{(1+X)^{\,p}-1}{X}\,,$$ clearly a monic polynomial of degree $p-1$, constant term $p$, all intermediate coefficients divisible by $p$. Now we have \begin{align} \Phi_{p^{n+1}}(X)&=\frac{X^{p^{n+1}}-1}{X^{p^n}-1}\\ \Phi_{p^{n+1}}(1+X)&=\frac{(1+X)^{p^{n+1}}-1}{(1+X)^{p^n}-1}\\ &=g\bigl[(1+X)^{p^n}-1\bigr]\,, \end{align} where what’s getting substituted into $g$ is a polynomial of degree $p^n$ without constant term, and in which all the terms but the highest one are divisible by $p$. That is, the last line can be written $g\bigl[pXh(X)+X^{p^n}\bigr]$, and since $g$ is Eisenstein, so is the last line of the display.