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I am really struggling to show that. I can't find a proof for $f$ to be irreducible. Eisentsien doesn't work. Revesing doesn't lead me anywhere and mod p didn't work as well, is there any criterion I might be missing? I have all the roots of $f$, but I don't think that is very useful for thir step.

Rolando Teixeira
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There is a general naive algorithm for testing the irreducibility of integer polynomials (which also works over number fields).

If $f$ is not irreducible then there is $g\in \Bbb{Z}[t]$ monic of degree $\le 4$ dividing it and whose roots have absolute value $\le 3^{1/4}$.

This implies that the coefficients of $g$ are smaller than those of $(t+3^{1/4})^{\deg(g)}$.

Just try the finitely many such polynomials and see if they divide $f$.

reuns
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  • I don't think i get it. Where does that come from, is that following gauss' Lemma? Thats the most similar thing i can find in my notes. I very much appreciate the help as I've been stuck on this for hours, and need to complete this one to proceed to the next part of the exercise. – Rolando Teixeira May 15 '22 at 18:39
  • Also, for the same exercise i also need to determine $|\mathbb{Q}(\alpha): \mathbb{Q} |$ is there any easier way to do this without know that $f$ is the minimal polynomial? If so I could use that to trivially prove the fact that $f$ is the minimal polynomial right? – Rolando Teixeira May 15 '22 at 18:44
  • Yes Gauss lemma says that $g$ must be a monic integer polynomial. You prove that $f$ is irreducible and you automatically get that $[\Bbb{Q}(\alpha):\Bbb{Q}]=\deg f$. The other way around doesn't have to be obvious (showing that $i+\sqrt2$ doesn't have a square root in $\Bbb{Q}(i,\sqrt2)$) – reuns May 15 '22 at 18:46