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I'm looking for a Galois extension $F$ of $\mathbb{Q}$ whose associated Galois group $\mbox{Gal}(F, \mathbb{Q})$ is isomorphic to $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. I'm wondering if I should just consider some $u \notin Q$ whose minimum polynomial in $Q[x]$ has degree 9. In that case we'd have $[\mathbb{Q}(u) : \mathbb{Q}] = 9$ and thus $|\mbox{Gal}(\mathbb{Q}(u), \mathbb{Q})| = 9$ (assuming the extension is Galois). But since there are two distinct groups of order 9 (up to isomorphism), I'm not sure that this will yield the desired result.

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It turns out that if you pick a polynomial at random, it's unlikely that the extension will be Galois (even if you could guarantee what it's Galois group would be). This can be quantified in various ways.

if $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are linearly disjoint, Galois extensions, then the Galois group of the compositum $LK$ is equal to the product of the Galois groups. So we're reduced to finding linearly disjoint Galois extensions of degree $3$. (Conversely, all extensions with your Galois group arise this way, by Galois theory).

Since these cubics are Galois and $3$ is prime, "linearly disjoint" is the same as "intersection equals $\mathbb{Q}$. A cubic polynomial generates a cyclic cubic field if and only if its discriminant is a square, so you can write down any two irreducible cubic polynomials with coprime square discriminant, for example. For more on cubic extensions, this is nice:

http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf

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  • $\begingroup$ Are you aware of a simple proof that if $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are linearly disjoint, then the Galois group of the compositum $LK$ is equal to the product of the Galois groups? $\endgroup$ – Krishnamurti2843 Feb 23 '18 at 5:15
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    $\begingroup$ Gotta love Keith Conrad. @Krishnamurti2843, I have discussion on the matter here. $\endgroup$ – Kaj Hansen Feb 23 '18 at 6:50
  • $\begingroup$ @Krishnamurti2843. In your situation, once we know $LK$ is Galois over $\mathbb{Q}$, it's not hard: there is a natural map from $\text{Gal}(LK/\mathbb{Q})$ to $\text{Gal}(L/\mathbb{Q}) \times \text{Gal}(K/\mathbb{Q})$, which is obviously injective (if an aut is the identity on $L$ and $K$ it's the identity on $LK$) and now we win -- it must also be surjective, by counting degrees. The fact that linear disjointness implies that the compositum is Galois takes a little more work but is not needed to solve your particular problem. $\endgroup$ – hunter Feb 24 '18 at 5:05
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This is something that nowadays can be easily done by using the LMFDB Database.

For example, if you go to the above link, on the left hand side you'll find a column with the words Number Fields and the options Global and Local. By clicking on the Global option you'll be taken to a page that allows you to search for number fields, i.e., finite extensions of $\mathbb{Q}$, and in particular it allows you to do a refined search according to different parameters that you might be interested in.

In this case, you can restrict the degree of your number fields to be 9 and you can choose the Galois group to be $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z}$ by typing the option C3xC3.

For instance, a simple search with those options easily shows that if $\alpha$ is a root of the polynomial $f(x):= x^9 - 15x^7 - 4x^6 + 54x^5 + 12x^4 - 38x^3 - 9x^2 + 6x + 1$, then the number field $F = \mathbb{Q}(\alpha)$ is Galois over $\mathbb{Q}$ and has Galois group $\mathrm{Gal}(F/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z}$.

I encourage you to try to learn to use the website, since it provides a wonderful tool for doing experiments in number theory and in related areas.

PS: As a useful exercise, why don't you try changing the condition on the Galois group to the other possible group of order 9. That will definitely answer your question about whether just choosing a root of an irreducible polynomial of degree 9 might work ;-)

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    $\begingroup$ First PIN number, then ATM machine, and now LMFDB database. ;) $\endgroup$ – André 3000 Feb 23 '18 at 5:04
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Take $\alpha = \exp(2\pi i/7),\ \beta = \exp(2\pi i/13)$, primitive roots of unity of order 7 and 13 (these are prime numbers and leave a remainder 1 when divided by 3). They generate fields having cyclic Galois groups of order 6 and 12 respectively. Each of them have their own (unique) cubic subfields having cyclic group of order 3 as their Galois groups.

So my guess is the following algebraic number should generate the field you want: $\alpha +\bar \alpha +\beta +\beta^3+\beta^9$

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$F=\Bbb Q(\cos(2\pi/7),\cos(2\pi/9))$ works.

It's the compositum of two linear disjoint Galois cubic extensions.

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  • $\begingroup$ How did you find $\cos(2\pi/7)$ and $\cos(2\pi/9)$? $\endgroup$ – Krishnamurti2843 Feb 23 '18 at 5:21
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    $\begingroup$ @Krishnamurti2843 from cyclotomic fields, as in P Vanchinathan's answer, but with slightly simpler numbers :-) $\endgroup$ – Lord Shark the Unknown Feb 23 '18 at 5:35
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http://www.math.ku.edu/~dlk/abgal.pdf

This paper might help. It solves the inverse Galois problem for all abelian groups, and your group is abelian.

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    $\begingroup$ It is quite unlikely that the easiest route to solve this problem be to solve it for all abelian groups! $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '18 at 4:25

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