# Find a polynomial over $\mathbb{Q}$ with a given Galois group.

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 $\mathbb{Q}$.

• Welcome to Mathematics Stack Exchange, due to site policy, could you show your current working – Alex Robinson Apr 24 '17 at 17:00
• Please see math.meta.stackexchange.com/questions/588/… for information on how to attract quality answers. – mlc Apr 24 '17 at 17:20
• This is a hard problem in general. See en.wikipedia.org/wiki/Inverse_Galois_problem – lhf Apr 24 '17 at 17:36
• $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\zeta_3)/\mathbb{Q}(\zeta_3)$ has $\mathbb{Z}_3\times \mathbb{Z}_3$ as its Galois group ? – reuns Apr 24 '17 at 17:51
• "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. – Dietrich Burde Apr 24 '17 at 18:08

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 $\mathbf Q$.

• This is an excellent answer! – Stella Biderman Apr 24 '17 at 19:53