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I came across the problem to show that when $n,m\in \mathbb{N}$ are coprime then the polynomial $X^{nm}-2^n3^m$ is irreducible over $\mathbb{Z}$. I solved it appealing to knowledge of complex numbers, that is, showing that if $\alpha \in \mathbb{C}$ is a root of such polynomial then $|\alpha |=2^{1/m}3^{1/n}$ and after showing that $|\alpha |^k\notin \mathbb{Z}$ for any $k\in\{1,\ldots,nm-1\}$, what makes impossible to write such polynomial as the product of two polynomials of $\mathbb{Z}[X]$.

However I suppose that there must be a more elegant argument using some number theory and some irreducibility criterion as reducing the polynomial to some $\mathbb{Z}_p$ field for appropiate prime $p$. Anyway Im just supposing that such argument must exists.

To resume, I opened this question to see if someone have a more purely algebraic or number theoretic argument to show the irreducibility of such polynomial over $\mathbb{Z}$.

Masacroso
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    in general , it is far from trivial to show that some polynomial is irreducible over $\mathbb Q[x]$. Eisenstein deos here only work if $m=1$ or $n=1$. I doubt that there is something better than the proof you have shown. – Peter Jul 07 '23 at 06:53
  • @Peter thank you for your comment, I'm starting to learn field theory so I suppose that there must exists other arguments. – Masacroso Jul 07 '23 at 06:54
  • This is a special case of When is $X^n-a$ is irreducible over F? – Sil Jul 07 '23 at 07:00
  • @Sil But the linked question is about irreducibility over a field. Here we care about irreducibility over a ring. – Vercassivelaunos Jul 07 '23 at 07:06
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    @Vercassivelaunos By Gauss lemma irreducibility is equivalent to irreducibility over rationals which is a field – Sil Jul 07 '23 at 07:07
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    The polynomial is primitive (coprime coefficients) , hence irreducibility over $\mathbb Z[x]$ implies irreducibility over $\mathbb Q[x]$ – Peter Jul 07 '23 at 07:10

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