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How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?

Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2\sqrt{3} = (\sqrt{3} - \sqrt{2}) + (\sqrt{2} + \sqrt{3})$ etc.

But how can i express $\sqrt{2}$ or $\sqrt[3]{3}$ with $\sqrt{2} + \sqrt[3]{3}$?

Is $\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ equal to $\mathbb Q[\sqrt{2}, \sqrt[3]{3}]$?

Thanks

Ilya Fedorov
  • 171

2 Answers2

2

Hint: Check which automorphisms of ${\mathbf Q}[\sqrt 2,\sqrt[3]{3},e^{2\pi i/3}]$ fix ${\mathbf Q}[\sqrt2+\sqrt[3]{3}]$.

tomasz
  • 35,474
0

Hint: It is enough to prove that $\mathbb Q[\sqrt{2} + \sqrt[3]{3}] =\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ because then the facts below imply that the degree of $\sqrt{2} + \sqrt[3]{3}$ is exactly $6$:

  • $\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ has degree at least $6$

  • $\sqrt{2} + \sqrt[3]{3}$ is a root of a polynomial of degree $6$

  • $\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ has degree at most $6$

Alternatively, $\sqrt{2} + \sqrt[3]{3}$ is a root of $x^6 - 6 x^4 - 6 x^3 + 12 x^2 - 36 x + 1$, which is irreducible because it is irreducible mod $13$. But that's a lot of work to do by hand.

lhf
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