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How can I find a polynomial over $\mathbb{Q}$ given its Galois group over $\mathbb{Q}$.

Do I have to guess, i.e. pick random polynomials and calculate their Galois groups, or is there a clever method?

I'm asking this, because I have an assignment that asks to find a polynomial over $\mathbb{Q}$, such that its Galois group is $\mathbb{Z}_3\times\mathbb{Z}_3$ over $\mathbb{Q}$.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange, due to site policy, could you show your current working $\endgroup$ – Alex Robinson Apr 24 '17 at 17:00
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    $\begingroup$ Please see math.meta.stackexchange.com/questions/588/… for information on how to attract quality answers. $\endgroup$ – mlc Apr 24 '17 at 17:20
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    $\begingroup$ This is a hard problem in general. See en.wikipedia.org/wiki/Inverse_Galois_problem $\endgroup$ – lhf Apr 24 '17 at 17:36
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    $\begingroup$ $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\zeta_3)/\mathbb{Q}(\zeta_3)$ has $\mathbb{Z}_3\times \mathbb{Z}_3$ as its Galois group ? $\endgroup$ – reuns Apr 24 '17 at 17:51
  • $\begingroup$ "I' am asking this, because I have an assignment" - note, that the assignment asks for something much easier. Like you would ask how to find all twin primes, because the assignment would ask to show that $(5,7)$ is a twin prime pair. $\endgroup$ – Dietrich Burde Apr 24 '17 at 18:08
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For a given finite abelian group $ G $, there is a relatively straightforward algorithm which produces a Galois extension $ L/\mathbf Q $ with Galois group $ G $. Then, the minimal polynomial of any primitive element of this extension has Galois group $ G $ over $ \mathbf Q $.

For this algorithm, use the structure theorem for finite abelian groups to write

$$ G \cong C_{n_1} \times C_{n_2} \times \ldots \times C_{n_k} $$

for some $ n_i $. Now, find distinct prime numbers $ p_1, p_2, \ldots, p_k $ such that $ p_i \equiv 1 \pmod{n_i} $ for each $ i $. (Such primes always exist by Dirichlet's theorem on arithmetic progressions.) Let $ d = p_1 p_2 \ldots p_k $, and let $ \zeta $ be a primitive $ d $th root of unity. Then, we have that

$$ \textrm{Gal}(\mathbf Q(\zeta_d)/\mathbf Q) \cong (\mathbf Z/d\mathbf Z)^{\times} \cong \prod_{i=1}^k (\mathbf Z/p_i \mathbf Z)^{\times} \cong \prod_{i=1}^k C_{p_i - 1} $$

Now, quotienting this group by the subgroup

$$ H = \prod_{i=1}^k C_{(p_i - 1)/n_i} $$

gives the group $ G $, and quotienting by $ H $ is equivalent to finding the Galois group of the fixed field of $ H $ over $ \mathbf Q $. Therefore, the fixed field of $ H $ in $ \mathbf Q(\zeta_d) $ is a Galois extension of $ \mathbf Q $ with Galois group $ G $.

Let's see how this works in practice. We want to find a Galois extension of $ \mathbf Q $ with Galois group $ C_3 \times C_3 $, so we want two distinct primes which are both $ 1 $ modulo $ 3 $. We may choose $ p_1 = 7 $ and $ p_2 = 13 $, so that $ d = 91 $. The fixed field of $ H $ is then the compositum of the cubic subfields of $ \mathbf Q(\zeta_7) $ and $ \mathbf Q(\zeta_{13}) $. With some help from Wolfram and the normal basis theorem, we may find that these cubic subfields are the splitting fields of $ X^3 + X^2 - 2X - 1 $ and $ X^3 + X^2 - 4X + 1 $, respectively. Therefore, the product

$$ (X^3 + X^2 - 2X - 1)(X^3 + X^2 - 4X + 1) $$

is a polynomial with Galois group $ C_3 \times C_3 $ over $ \mathbf Q $.

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    $\begingroup$ This is an excellent answer! $\endgroup$ – Stella Biderman Apr 24 '17 at 19:53

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