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Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$

My attempt:

Firstly, since, $\Bbb Q(\sqrt{2}+3^{1/3}) \subseteq \Bbb Q(\sqrt{2},3^{1/3})$ , I computed $(\sqrt{2}+3^{1/3})^{-1} = 6-4\sqrt{2}+4(3^{1/3})+3(3^{2/3})-3(\sqrt{2}3^{1/3})-2(\sqrt{2}3^{2/3})$ and also tried some naive manipulation by considering powers of $(\sqrt{2}+3^{1/3})$ and tried to eliminate terms ( imitating the proof for $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2}+\sqrt{3})$ ) but it gets very messy.

Secondly, I am trying this problem by Galois theory but since, $\Bbb Q(\sqrt{2},3^{1/3})|\Bbb Q$ is not Galois,I don't get how to proceed.

Thanks in advance for help!

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  • $\begingroup$ See here for a strategy applied to a related problem. $\endgroup$ – Kaj Hansen Mar 17 at 8:12
  • $\begingroup$ @KajHansen But the problem over here (as mentioned in the question too) $\Bbb Q(\sqrt{2}, 3^{1/3})|\Bbb Q$ is not a Galois extension unlike $\Bbb Q(2^{1/4},i) | \Bbb Q$ $\endgroup$ – reflexive Mar 17 at 8:18
  • $\begingroup$ Ah yes, you are quite right $\endgroup$ – Kaj Hansen Mar 17 at 20:18
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Let's use Galois theory. You surely already know $K=\Bbb Q(\sqrt2,\sqrt[3]3)$ has degree $6$ over $\Bbb Q$. Therefore its Galois closure $L=\Bbb Q(\sqrt2,\sqrt[3]3,\omega)$ where $\omega=\exp(2\pi i/3)$ has degree $12$. The Galois group $G$ has order $12$ and is isomorphic to $S_2\times S_3$: the $S_2$ swaps $\pm\sqrt2$ and the $S_3$ permutes the $\omega^k\sqrt[3]3$. The only non-trivial element of $G$ fixing $\sqrt2+\sqrt[3]3$ is the complex conjugation. Therefore $\sqrt2+\sqrt[3]3$ has six distinct conjugates, and generates a field of degree $6$, which can only be $K$.

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Let $\alpha = \sqrt2+ \sqrt[3]{3}$.

First isolate $\sqrt[3]{3}$ and raise everything to the third power:

$$\sqrt[3]{3} = \alpha - \sqrt{2} \implies 3 = (\alpha - \sqrt{2})^3 = \alpha^3-3\alpha^2\sqrt{2}+6\alpha - 2\sqrt{2}$$ Then isolate $\sqrt{2}$ $$\sqrt{2}(3\alpha^2+2) = \alpha^3+6\alpha-3$$ so $$\sqrt{2} = \frac{\alpha^3+6\alpha-3}{3\alpha^2+2} \in \mathbb{Q}(\alpha)$$ Then also $\sqrt[3]{3} = \alpha - \sqrt{2} \in \mathbb{Q}(\alpha)$ so $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt[3]{3})$.

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  • $\begingroup$ How do you know $\dfrac 1{3\alpha^2+2}\in\mathbb Q(\alpha)$ ? You need to prove either it is a polynomial in $\alpha$ like the numerator, or the ratio of both is a polynomial in $\alpha$. I feel like the claim is a bit hasty. $\endgroup$ – zwim Mar 17 at 22:26
  • $\begingroup$ @zwim Umm... we know that $\alpha \in \mathbb{Q}(\alpha)$, $\mathbb{Q} \subseteq \mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha)$ is closed under field operations. Therefore $$\alpha \in \mathbb{Q}(\alpha) \implies \alpha^2 \in \mathbb{Q}(\alpha) \implies 3\alpha^2 \in \mathbb{Q}(\alpha) \implies 3\alpha^2 +2 \in \mathbb{Q}(\alpha) \implies \frac1{3\alpha^2+2} \in \mathbb{Q}(\alpha)$$ $\endgroup$ – mechanodroid Mar 17 at 22:30
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There are two general theorems which could be useful in such situations as yours :

Thm.1 : Let $N$ be the splitting field of a separable irreducible polynomial $f$ having prime degree $p\neq char(K)$, and let $a_1, a_2$ two distinct roots of $N$. Then $N$ is a normal closure of $K(a_1 - a_2)$. Proof : Let $M$ be the normal closure of $K(a_1 - a_2)$ in $N$. By construction, $Gal(N/K)$ contains an automorphism $\sigma$ which is a $p$-cycle on the $a_i$, say $\sigma = (a_1, a_2,...,a_p)$ with an adequate indexation. As $\sigma$ stabilizes $M$, all the differences $a_1 - a_2, a_2 - a_3,..., a_{p-1} - a_p$ lie in $M$, hence all the $a_1 - a_j$ lie in $M$. Adding up, one gets that $pa_1 - (a_1 +a_2 +...+a_p)\in M$. Since the later sum lies in $K$ and $p\neq char(K)$, it follows that $M=N$.

Thm.2 : Let $a, b$ be two non null elements separable over $K$, s.t. the degree of $a$ over $K$ is a prime $p\neq char(K)$, and the degree of $b$ is $n<p$. Then $K(a,b)=K(a+b)$. Proof : Let $M, N$ be the normal closures of $K(a), K(b)$ in a normal closure of $K(a,b)$. Let us show that $a_i - a_j \neq b_k - b_l$ for all pairs $(i,j)$ and $(k,l)$. Indeed, if $a_i - a_j = b_k - b_l$, then $a_i - a_j \in N$, and thm.1 implies that $M\subset N$ : impossible because $[N:K]$ divides $n!$ and hence is not divisible by $p$.

In your case, one can apply thm.2 to $a=\sqrt [3] 3$ and $b=\sqrt 2$ .

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