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I am trying to find the minimal polynomials of

1) $\alpha = \sqrt[4]{2}$ on the field $\mathbb{Q}$

2) $\alpha = \sqrt[4]{2}$ on the field $\mathbb{Q}(\sqrt{2})$

3) $\alpha = \sqrt{2} + \sqrt[3]{2}$ on $\mathbb{Q}$

I have no idea how to go about and I think the solution to these would help me to understand

Here's my attempt for 1)

$x=\sqrt[4]{2} \implies x^4-2=0$ is the minimum polynomial. Is that the right way to approach it?

Sylvester Stallone
  • 937
  • yes, that is the right way – Saketh Malyala Jun 19 '17 at 04:22
  • You should prove that $x^4 - 2$ is minimal, i.e. that this polynomial is irreducible over $\mathbb{Q}$. Eisenstein's criterion may help with $1$ and $3$. For #$2$, there are two options -- either the minimal polynomial is quadratic or linear. Think of what polynomial over $\mathbb{Q}(\sqrt{2})$ naturally has $\sqrt[4]{2}$ as a root, and prove the linear case is impossible. – Santana Afton Jun 19 '17 at 04:56
  • Related. I am fairly sure that we have the calculations for 3) carried out exactly here. When/if somebody finds it, we can close this is a dupe. – Jyrki Lahtonen Jun 19 '17 at 05:26

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