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I need to find degree of minimal polynomial $20^{\frac{1}{3}} + 5^{\frac{1}{3}}$ over $\mathbb{Q}\left(i\sqrt{3}\right)$. I know that over $\mathbb{Q}$ this polynomial has degree $9$. Also, I know that this element lies in $\mathbb{Q}(20^{\frac{1}{3}},5^{\frac{1}{3}})$

Any hints?

egorovik
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chaseperfection
  • 97
  • What is the Galois group of $\mathbb{Q}(2^{\frac{1}{3}},5^{\frac{1}{3}},\zeta_3)/\Bbb{Q}(\zeta_3)$ – reuns Dec 15 '19 at 15:25
  • $$\mathbb{Q}(20^{\frac{1}{3}},5^{\frac{1}{3}})=\mathbb{Q}(20^{\frac{1}{3}}+5^{\frac{1}{3}})$$ – Angina Seng Dec 15 '19 at 15:26
  • @reuns i have computed $\mathbb{Q}(2^(\frac{1}{3}),5^(\frac{1}{3}),3-unity)$ and its a group of order 18. But how to take quotient – chaseperfection Dec 15 '19 at 18:12
  • $G=Gal(\mathbb{Q}(2^{\frac{1}{3}},5^{\frac{1}{3}},\zeta_3)/\Bbb{Q}(\zeta_3))$ is $C_3\times C_3$. Then the $\Bbb{Q}(\zeta_3)$-minimpal polynomial of $\alpha=20^{\frac{1}{3}}+5^{\frac{1}{3}}$ is $\prod_{\beta\in G(\alpha)} (x-\beta)$, you can decompose it as a product of 3 cubic polynomials. – reuns Dec 15 '19 at 18:18
  • @reuns but why it is $C_{3} \times C_{3}$ – chaseperfection Dec 15 '19 at 18:23
  • Because $\mathbb{Q}(2^{\frac{1}{3}},\zeta_3)/\Bbb{Q}(\zeta_3),\mathbb{Q}(5^{\frac{1}{3}},\zeta_3)/\Bbb{Q}(\zeta_3)$ are two degree $3$ Galois extensions with trivial intersection. – reuns Dec 15 '19 at 18:36
  • See e.g. https://math.stackexchange.com/a/3477323/300700 – nguyen quang do Dec 15 '19 at 20:05

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