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

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.

• See here for a strategy applied to a related problem. – Kaj Hansen Mar 17 at 8:12
• @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 $\Bbb Q(2^{1/4},i) | \Bbb Q$ – reflexive Mar 17 at 8:18
• Ah yes, you are quite right – Kaj Hansen Mar 17 at 20:18

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 $$K$$.

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 $$\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt[3]{3})$$.

• 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. – zwim Mar 17 at 22:26
• @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 $$\alpha \in \mathbb{Q}(\alpha) \implies \alpha^2 \in \mathbb{Q}(\alpha) \implies 3\alpha^2 \in \mathbb{Q}(\alpha) \implies 3\alpha^2 +2 \in \mathbb{Q}(\alpha) \implies \frac1{3\alpha^2+2} \in \mathbb{Q}(\alpha)$$ – mechanodroid Mar 17 at 22:30

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 $$n. Then $$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 $$b=\sqrt 2$$ .