Prove that if $x^{nm}- a$ is irreducible then $x^n-a$ and $x^m-a$ is irreducible

Asked Sep 01 '20 at 20:25

Active Sep 01 '20 at 20:53

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Let $a \neq 0$ and let $x^{nm}- a$ be a polynomial. Prove that if $x^{nm}- a$ is irreducible then $x^n-a$ and $x^m-a$ are irreducible.

$\textbf{My attempt:}$

Suppose that $x^{m}-a$ is reducible, then there exists $p,q$ s.t $x^m-a = p(x)q(x)$, with $\deg p $ and $\deg q<m$.

Now note that $x^{nm}-a = (x^n)^m - a = p(x^n)q(x^n)= r(x)s(x)$, then $x^{nm}-a$ is reducible.

My question is, $\deg r, \deg s < mn$??

Also, is that correct?

edited Sep 01 '20 at 20:53

Bernard

asked Sep 01 '20 at 20:25

Joãonani

2

I think this is good. Since $p(x)$ has degree strictly less than $m$, the degree of $p(x^n)$ must be strictly less than $mn$. Likewise with $q(x)$. – morrowmh Sep 01 '20 at 20:31
Specifically: $\deg r = n\deg p$ and $\deg s=n\deg q$. – TonyK Sep 01 '20 at 20:55

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