Constructive Inverse Galois Problem for finite abelian groups

Question

I have seen a solution to the Inverse Galois Problem for finite abelian groups. However, to gain a better understanding of what's going on, I want to use this argument on the group $\mathbb{Z} / 8 \mathbb{Z}$ in a more constructive way.

The general argument begins with the following:

$G := \mathbb{Z} / 8\mathbb{Z}$ is a subgroup of $\left(\mathbb{Z} / 17 \mathbb{Z} \right)^{\times}\cong \mathbb{Z} / 16 \mathbb{Z}$, so let $H := \left(\mathbb{Z} / 17 \mathbb{Z} \right)^{\times} / G \cong \mathbb{Z} / 2 \mathbb{Z}$, then by the Galois Correspondence, we have that: $$\text{Gal} \left(\mathbb{Q} (\zeta)^H / \mathbb{Q} \right) \cong \text{Gal} \left(\mathbb{Q} (\zeta) / \mathbb{Q} \right) / H \cong \left(\mathbb{Z} / 17 \mathbb{Z} \right)^{\times} / \mathbb{Z} / 2 \mathbb{Z} \cong G $$ where $\zeta$ is a 17th root of unity, and the notation $L^H$ denotes the subfield of $L$ fixed by automorphisms in $H$.

Next, $\mathbb{Z} / 2 \mathbb{Z} \subset \left(\mathbb{Z} / 17 \mathbb{Z} \right)^{\times} $ corresponds to the automorphisms $\left \{\textbf{id}, \theta^*: \zeta \rightarrow \zeta^{16} \right \}$, since $\theta^*$ is of order 2, since $16^2 \equiv -1^2 \mod 17 \equiv 1 \mod 17$. Finally, $$\mathbb{Q}(\zeta)^H/\mathbb{Q} = \{\alpha = b_0 + b_1 \zeta + ... + b_{16} \zeta^{16} : \theta^{*} (\alpha) = \alpha, \hspace{4pt} b_i \in \mathbb{Q} \} $$ I am pretty convinced that the requirement that $\theta^* (\alpha) = \alpha$ implies that $b_1 = b_{16}$, $b_2 = b_{15}$, ... , $b_8 = b_9$.

Is it safe to say that all that remains is to realize $\mathbb{Q}(\zeta)^H/\mathbb{Q}$ as the splitting field of some polynomial, this polynomial will then have Galois group $\mathbb{Z} / 8\mathbb{Z}$ ? Am I missing anything? If not, how would one realize this field as the splitting field of a polynomial?

Answer 1

Your use of the Galois correspondence is a bit off.

Let $L=\mathbb{Q}(\zeta)$ be the $17$th cyclotomic field, and let $\Gamma=\operatorname{Gal}(L/\mathbb{Q})\simeq \mathbb{Z}/16\mathbb{Z}$. If you want $G=\operatorname{Gal}(K/\mathbb{Q})$ for some $K=L^H\subset L$, you should take $H$ to be a subgroup and not a quotient of $\Gamma$, so that $G\simeq \Gamma/H$.

In your question the problem is that you see $G$ both as a subgroup and a quotient of $\Gamma$. Of course it is true up to isomorphism, but it is more or less a coincidence. If you define $H$ as a quotient of $\Gamma$, in general $L^H$ does not make sense (here it sort of works because there is a unique subgroup of $\Gamma$ isomorphic to $H$ so we understand what you mean, but in general it won't work).

So, once you define $H$ as the subgroup of $\Gamma$ such that $\Gamma/H\simeq G$, then indeed you have $G=\operatorname{Gal}(L^H/\mathbb{Q})$, which solves the inverse Galois problem for $G$. Finding a specific polynomial such that $L^H$ is its splitting field is a rather unrelated issue (but you can do it if you want!).