Irreducibilty Cyclotomic polynomial $f(x^{n})$

Question

Let $p$ be a prime.

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

Let $g(x)$ = $f(x^{n})$ where n is any positive integer.

I know $f(x)$ is irreducible by Eisenstein's criterion.

Now i want to show $g(x)$ is irreducible:

If $g(x) = h(x)\cdot k(x) $,

Then $h(x) = (x^{n})^{k} + (x^{n})^{k-1} +....+1$,

And $k(x) = (x^{n})^{l} + (x^{n})^{l-1} +...+ 1$.

If any terms of $k(x)$ or$ h(x)$ is not of the form of $(x^{n})^{m}$ then when multiplied with 1 produces terms which is not in $ g(x).$ Then we can substitute $ x^{n}$ as $ y $ Which will produce that cyclotomic polynomial $ f(x)$ is reducible.

Which is a contradiction

Is my proof correct?

Answer 1

If $p |n$ then $\Phi_n(x^{p^k})$ is irreducible. This is because $\zeta_{np^k}$ is a root of $\Phi_n(x^{p^k})$ ie. $\Phi_{np^k}|\Phi_n(x^{p^k})$, and we know that $\deg \Phi_{np^k} = \varphi(np^k)=p^k \varphi(n)=\deg(\Phi_n(x^{p^k}))$

If $p\nmid n$ then $\deg \Phi_{np^k} = \varphi(np^k)=(p-1)p^{k-1} \varphi(n)$ so $\Phi_n(x^{p^k})$ isn't irreducible. Example $\Phi_2(x^3)=x^6+1$.

For the same reason $\Phi_n(x^m)$ is irreducible iff $\varphi(nm)=m\varphi(n)$ iff each prime divisor of $m$ divides $n$.