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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 $\mathbb Q(\sqrt[3]{2},\sqrt[3]{3})$?

Jyrki Lahtonen
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Isaac Lou
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    Most probably, you can take $\alpha = \sqrt[3]{2}+\sqrt[3]{3}$ – lhf Jun 20 '20 at 19:23
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    by a simple geometric-probabilistic argument, you can see that almost any linear combination of the separate generators will be a field generator of the big field. – Lubin Jun 20 '20 at 19:24
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    Since you know that $[\mathbb Q(\sqrt[3]{2},\sqrt[3]{3}):\mathbb Q]=9$ it is enough to prove that $x^9 - 15 x^6 - 87 x^3 - 125$ is irreducible because $\alpha = \sqrt[3]{2}+\sqrt[3]{3}$ is root. – lhf Jun 20 '20 at 19:29
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    I can not see yet that your polynomial is irreducible, @lhf . Can you? – Lubin Jun 21 '20 at 03:02
  • @Lubin, nor can I. – lhf Jun 21 '20 at 12:20
  • Adapt the solution in https://math.stackexchange.com/questions/3666493/field-extension-of-degree-three – lhf Jun 21 '20 at 12:23
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    Ha ! A method I never would have guessed. Excellent, @lhf . – Lubin Jun 21 '20 at 14:04
  • Put $k=Q(\zeta_3)$, so that $K=k(\sqrt [3] 2, \sqrt [3] 3)$ is galois, with group $G=C_3 \times C_3$. To determine the cubic subextensions of $K/k$ without too many computations, just notice that the associated Kummer radical $R$ (= dual of $G$) is a subgroup of $k^/{k^}^3$ of order $9$, generated by the classes [2] and [3] mod ${k^*}^3$. In additive notation, $R$ can be viewed as an $\mathbf F_3$ - vector space of dimension $2$, and then the cubic subextensions correspond bijectively to the $\mathbf F_3$ - lines contained in $R$. Such a line contains exactly two non null vectors... – nguyen quang do Jun 21 '20 at 15:36
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    so the number of lines is $(9-1)/(3-1)=4$. It follows that the cubic subextensions are exactly the fields $k(\sqrt [3] a)$ for $a=2,3,6,12$ ($12$ and $18$ give rise to the same extension because their product is a cube), which are obviously distinct from $k(\sqrt [3] 2+ \sqrt [3] 3)$ (by Kummer again). Note that this approach works for any prime in place of $3$. Anyway it dispenses us with consideration of irreducible polynomials. – nguyen quang do Jun 21 '20 at 15:56
  • Good, @nguyenquangdo . I thought of that method, did not carry it through. – Lubin Jun 21 '20 at 17:35
  • Quite a bit of information about the intermediate fields here. As long as the generator does not belong to any of them, you are fine as nguyen quang do explained. – Jyrki Lahtonen Jun 21 '20 at 20:30

2 Answers2

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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.

Jyrki Lahtonen
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    Moral: Finding primitive elements becomes trivial after you have figured out the Galois group. The exercise leaves scope for ingenuity when developing the theory, and many useful tools (characteristic polynomials of suitable linear transformations and such) can/should be introduced along the way. In the end the task becomes somewhat boring. – Jyrki Lahtonen Jun 22 '20 at 06:10
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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.