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I know that

$$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]=[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]\cdot[\mathbb{Q}(3^{1/4}):\mathbb{Q}],$$ and

$$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]=[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot[\mathbb{Q}(2^{1/3}):\mathbb{Q}].$$

But $[\mathbb{Q}(3^{1/4}):\mathbb{Q}]=4$, and $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$

From here I can deduce that $lcm(4,3)=12$ divides $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$. This is, $12 \leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$.

I would like to show that the other inequality also holds, but I do not know how to continue here.

Another approach: I was triying to find the minimal polynomial of $2^{1/3}$ over $\mathbb{Q}(3^{1/4})$, or the minimal polynomial of $3^{1/4}$ over $\mathbb{Q}(2^{1/3})$ but I do not know how to justify that $x^3-2$ is irreducible over $\mathbb{Q}(3^{1/4})$, or that $x^4-3$ is irreducible over $\mathbb{Q}(2^{1/3})$.

Thanks for any help.

Jyrki Lahtonen
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Camilo Diaz
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    You can assume that $x^3-2 = (x+a)(x^2+bx+c)$, where $a,b,c \in \mathbb{Q}(3^{1/4})$ and you'll find that $a^3 = -2$. Then try to show that $a \not \in \mathbb{Q}(3^{1/4})$. – Ubik Jun 10 '24 at 09:27
  • @RHspqr If $-2^{1/3} \in \mathbb{Q}(3^{1/4})$, then $2^{1/3} \in \mathbb{Q}(3^{1/4})$. Thus, $\mathbb{Q}(2^{1/3},3^{1/4})=\mathbb{Q}(3^{1/4})$. This implies that $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=[\mathbb{Q}(3^{1/4}):\mathbb{Q}].$ Which leads to $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot3=4.$ This is, $3|4$. Contradiction. Hence, $x^3-2$ is irreducible over $\mathbb{Q}(3^{1/4})$. Finally, $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]\leq3$, giving $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}] \leq 12$. Therefore the equality – Camilo Diaz Jun 11 '24 at 04:41
  • @RHspqr Is this reasoning correct?

    Also, I would like to ask how can be proved that certain polynomial is of the least degree? for example, knowing that $x^3-2$ is irreducible over $\mathbb{Q}(3^{1/4})$, I only get that the minimal polynomial of $2^{1/3}$ over $\mathbb{Q}(3^{1/4})$ has at most degree 3. Right?

    Thanks!

     – Camilo Diaz Jun 11 '24 at 04:50
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    Yes that's a correct reasoning. For your question about least degree, we know that $K(\alpha) = K[x]/m_{\alpha}(x)$, which means that $m_{\alpha}(x)$ generates a maximal ideal over $K[x]$. However, since $K[x]$ is a PID, we know that all prime elements are irreducible and all prime ideals are maximal. Hence we know that $m_{\alpha} = f$. – Ubik Jun 11 '24 at 08:47
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    We may also prove this from the definition of irreducible polynomial. By definition, the minimal polynomial divides all $f \in K[x]$ such that $f(\alpha) = 0$, thus $m_{\alpha} |f$. However, as $f$ is irreducible, $m_{\alpha} = 1 or f$. – Ubik Jun 11 '24 at 08:48

1 Answers1

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If $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$, then $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]<12$, against the fact that 12 divides $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$, so $3\leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]$.

On the other hand $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \leq \deg(x^3-2)= 3$.

This works in general for $\mathbb F(a,b)/\mathbb F$, as soon as the degrees $[\mathbb F(a):\mathbb F]$ and $[\mathbb F(b):\mathbb F]$ are coprime (see this post).

Desperado
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    Then it cannot happen $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$, but $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \geq 3$ leads to $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}] \geq 12$, which is the inequality I had found. Isn't it? – Camilo Diaz Jun 10 '24 at 23:56
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    $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$ leads to a contradiction to the inequality you proved, so $3\leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]$. On the other hand $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \leq \deg(x^3-2)= 3$, and you are done. – Desperado Jun 11 '24 at 08:20