How to express the field $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$ as a simple extension of $\mathbb Q$?

Question

I’ve found $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3},\zeta_3):\mathbb Q$ and the galois group $G=(C_3\times C_3)\rtimes C_2$ which has order 18. Now $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$ is a subfield where $[\mathbb Q(\sqrt[3]{2},\sqrt[3]{3}):\mathbb Q]=9$, how can I express $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$ as a simple extension of $\mathbb Q$? i.e. how can I find $\alpha$ s.t. $\mathbb Q(\alpha)$ isomorphic to $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$?

Answer 1

As the commenters suspected, we can use $\alpha=\root3\of2+\root3\of3$ as the generator.

All the hard work was done when you figured out the Galois group $G$. By its description from here we see that the conjugates of $\alpha$ are $$ \zeta_3^j\root3\of2+\zeta_3^k\root3\of3 $$ with $j,k\in\{0,1,2\}$ varying independently. It is trivial to check numerically that different pairs of exponents give distinct numbers. Therefore $\alpha$ has nine conjugates, and consequently its minimal polynomial over the rationals has degree $9$, and also $[\Bbb{Q}(\alpha):\Bbb{Q}]=9$.

But $\alpha\in\Bbb{Q}(\root3\of2,\root3\of3)$, a field of degree at most nine (exactly nine as a step of the derivation of the Galois group). We are done.

Answer 2

As an alternative without Galois theory, one can use the constructive proof of the primitive element theorem. Let $\alpha = \sqrt[3]2$ and $\beta = \sqrt[3]3$. An element of the form $$\alpha + c \beta$$ is a primitive element for that extension unless $$c = \frac{\alpha' - \alpha}{\beta - \beta'}$$ where $\alpha',\beta'$ are conjugates of $\alpha,\beta$ respectively (and $\beta' \neq \beta$, of course). Thus we see that the only ``bad'' choices of $c$ are $$-\frac{1-\zeta ^i}{1 - \zeta^j} \frac{\alpha}{\beta}$$ where $\zeta$ is a primitive $3$rd root of unity, $i=0,1,2$ and $j=1,2$. Since we are taking $c\in\mathbb Q$, it is a real number. As we can take $\alpha/\beta$ can be taken to be real, we can rule out some choices of $i,j$ depending on whether the fraction involving the roots of unity is real.

For it to not be obviously real, we must have $i\neq j$ and $i\neq 0$, so you can just check $i=1$ and $j=2$ (the only other choice is just the inverse of this) which is not real.

In particular, it is never real unless $i=0$ or $i=j$, but then in the latter case we are taking $c = \alpha/\beta = \sqrt[3]{2/3}$ which is obviously not rational. The former case is $c=0$.

So in fact any linear combination of that form works except for the obviously bad choice of $c=0$.

There are coarser and easier options available too - it is not hard to bound the complex norm of those ``bad'' $c$'s, and then you can just pick a rational number larger than that.

This argument is actually sort of implicit in the other answer - I think (but could be wrong) that actually checking that those linear combinations are distinct would pass through an argument sort of like this. The constructive primitive element theorem basically just states the only possible ways that a linear combination could fail to produce ``as many conjugates as possible'' and observes that there are just finitely many.