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I wanted to prove irreducibility of $f(X)=X^5 + 126X^4 + 57X^3 + 21X + 300$ over $\mathbb{Q}(\sqrt[19]{2})$.

When I saw this I realized Eisenstein proved it is irreducible over $\mathbb{Q}$ because $3$ divides every term and it's square does not divide $300$.

Clearly $x^{19}-2$ is the minimal polynomial of $\sqrt[19]{2}$ over $\mathbb{Q}$ and so $[\mathbb{Q}(\sqrt[19]{2}):\mathbb{Q}]=19$. If there were a root of $f$, say $\alpha\in \mathbb{Q}(\sqrt[19]{2}) $, we would have: $\mathbb{Q}\subset \mathbb{Q}(\alpha)\subset \mathbb{Q}(\sqrt[19]{2})$. Because $19$ is prime, we must have $\mathbb{Q}(\alpha)=\mathbb{Q}$ which is absurd as $f$ has no roots on $\mathbb{Q}$ or $\mathbb{Q}(\sqrt[19]{2})=\mathbb{Q}(\alpha)$.

But this is clearly absurd, because $\deg(\min(\mathbb{Q},\alpha))\leq 5$, but it should be $19$. Therefore $f$ really has no roots over $\mathbb{Q}(\sqrt[19]{2})$. However this doesn't really help to prove $f$ is irreducible, because we must consider all other possible non trivial decompositions...

Kadmos
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    It's a general result that if $f\in F[x]$ is an irreducible polynomial, and $K$ is a field extension of $K$ such that $[K:F]$ is coprime to $\deg(f)$ then $f$ is also irreducible over $K$. See here: https://math.stackexchange.com/questions/155319/an-irreducible-polynomial-of-degree-coprime-to-the-degree-of-an-extension-is-irr – Mark Oct 17 '23 at 00:29
  • Thanks! I guess we should close this as a duplicate... – Kadmos Oct 17 '23 at 00:33

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