# Find an element $\theta \in \mathbb{R}$ such that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) = \mathbb{Q}(\theta)$

Find an element $\theta \in \mathbb{R}$ such that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) = \mathbb{Q}(\theta)$

I must find one element $\theta$ that

$$\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subseteq \mathbb{Q}(\theta)$$ and

$$\mathbb{Q}(\theta)\subseteq \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$

For example, I know that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$, because the sum of these two elements must also be in the set. I don't know, however, if the inverse inclusion holds. For example, I must be able to form $\sqrt{2}$ and $\sqrt[3]{5}$ using $\sqrt{2}+\sqrt[3]{5}$ using only multiplications of $\sqrt{2}+\sqrt[3]{5}$ by itself, using its inverse, and summing with the results of the multiplications, right?

$(\sqrt{2}+\sqrt[3]{5})^2 = 2 + 2\sqrt{2}\sqrt[3]{5} + \sqrt[3]{5^2}$ which isn't helpful at all, so I think this is not the right candidate.

• Are you allowed to use a CAS (such as Maple or Mathematica)? Commented Sep 11, 2017 at 0:56

Take $\theta=\sqrt{2}{\sqrt[3]{5}}$

We have that $(\sqrt{2}\sqrt[3]{5})^2=2\sqrt[3]{5^2} \Rightarrow \sqrt[3]{5^2} \in \mathbb{Q}(\theta)$

Therefore $\sqrt{2}=\frac{(\sqrt{2}\sqrt[3]{5})\sqrt[3]{5^2}}{5} \in \mathbb{Q}(\theta) \Rightarrow \sqrt{2}\in \mathbb{Q}(\theta)$

Now with the same way we can prove that $\sqrt[3]{5} \in \mathbb{Q}(\theta)$ by noticing that $\sqrt[3]{5}=\frac{\sqrt{2}(\sqrt{2}\sqrt[3]{5})}{2}$

Thus $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subseteq \mathbb{Q}(\theta)$

Also it is obvious that $\mathbb{Q}(\theta) \subseteq \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$

So you have the result.

• Very nice!${}{}{}$ Commented Sep 11, 2017 at 1:14
• I like this because it’s very elegant, and not the sort of trick I would have thought of. Plus one. Commented Sep 11, 2017 at 1:15

For your candidate $\theta =\sqrt{2} + \sqrt[3]{5}$ we have $$(\theta -\sqrt{2})^3=5\\ \theta^3 + 6\theta-(3 \theta^2 +2)\sqrt{2} =5\\ \sqrt{2}=\frac{\theta^3 + 6\theta - 5}{3\theta^2 +2}$$ and then use $\sqrt[3]{5}= \theta -\sqrt{2}$

• Most excellent. And since $\theta$ is real, $3\theta^2+2\ne0$. Commented Sep 11, 2017 at 13:46
• @Lubin: Thanks! Also, good point. Commented Sep 11, 2017 at 14:13

It is certainly true that your guess of $\theta=\sqrt2+\sqrt[3]5$ would work, and there are very many methods of showing this, though none is as quick as using instead the product of $\sqrt2$ and $\sqrt[3]5$ as suggested by @Marios.

For instance, consider the field $k=\Bbb Q(\sqrt2\,)$,whose ring of integers is $\Bbb Z[\sqrt2\,]$, known (or easily shown) to be a Principal Ideal Domain. In this ring, it happens that $5$ is still prime, though even if $5$ were to factor as $5=\pi\pi'$, my argument below would still apply:

Since $5$ is square-free in $\Bbb Z[\sqrt2\,]$, we then know that $\sqrt[3]5\notin k$, and as a result $\sqrt2+\sqrt[3]5\notin k$, and this quantity therefore generates the field $k(\sqrt[3]5\,)=\Bbb Q(\sqrt2,\sqrt[3]5\,)$.

• I follow the logic up to and including $$k(\sqrt{2}+\sqrt[3]{5}) =k(\sqrt[3]{5}) =\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ but I think further reasoning is needed to support the claim $$\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ Commented Sep 11, 2017 at 3:12
• You’re quite right, @quasi. I did not explain why $k\subset\Bbb Q(\sqrt2+\sqrt[3]5\,)$, a serious defect. I think orangeskid’s argument will work very nicely for this. Commented Sep 11, 2017 at 13:43