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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?

jonan
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    Yes. This works. You may also realize that the order two automorphism you found is (the restriction of) the usual complex conjugation. So the degree eight field consists of the real numbers of $\Bbb{Q}(\zeta)$. It is generated by $\zeta+\overline{\zeta}=2\cos(2\pi/17)$. It is a normal extension of $\Bbb{Q}$ because it is the fixed field of a normal subgroup of $(\Bbb{Z}/17\Bbb{Z})^*$. That part is kinda trivial because the big group is abelian, so all its subgroups are normal. – Jyrki Lahtonen Apr 02 '20 at 19:10
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    But, as Captain Lama (+1) explains, the early part is a bit strange. The point is that you want a subgroup $H$ such that $G/H$ is cyclic of order eight. Anyway, you correctly figured out that you want the fixed field of the unique subgroup of order two. – Jyrki Lahtonen Apr 02 '20 at 19:12
  • @JyrkiLahtonen thanks, I should have noticed the complex conjugation observation. Can I conclude that $\mathbb{Q}(\zeta)^{H} / \mathbb{Q} = { b_0 + b_1 \cos \left(\frac{2\pi}{17} \right) } + ... + b_8 \cos \left( \frac{16 \pi}{17} \right) : b_i \in \mathbb{Q} }$ ? I don't see why you have included the factor of 2 when you say $2\cos(2\pi/17)$ generates it. – jonan Apr 02 '20 at 20:15
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    Yes. The factor $2$ is unnecessary for this purpose. In algebraic number theory we work with so called algebraic integers. There the factor two is necessary, so I included it out of inertia. – Jyrki Lahtonen Apr 03 '20 at 05:09
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    If you want to find the minimal polynomial of $2\cos(2\pi/17)$, you can write the powers of $\zeta$ as follows: Instead of $\zeta^{16}$ use $\zeta^{-1}$, instead of $\zeta^{15}$ use $\zeta^{-2}$, ... instead of $\zeta^9$ use $\zeta^{-8}$. This way you can write all the nontrivial powers with exponents in the range $[-8,8]$. Then you can rewrite the equation $$0=\zeta^8+\zeta^7+\cdots+\zeta+1+\zeta^{-1}+\cdots+\zeta^{-7}+\zeta^{-8}$$ using the variable $u=(\zeta+\zeta^{-1})$, which is exactly what you want. – Jyrki Lahtonen Apr 03 '20 at 05:15
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    See this old answer of mine for an explanation, an example, and a generalization. Observe that with this technique you naturally get the minimal polynomial of $2\cos(2\pi/17)$. The key being that the (monic) minimal polynomial of the doubled cosine has integer coefficients (that's why the doubled cosine is an algebraic integer). If you transform it to the minimal polynomial of $\cos(2\pi/17)$, you get coefficients with powers of two in the denominator. – Jyrki Lahtonen Apr 03 '20 at 05:16

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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!).

Captain Lama
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  • Okay, so replacing $\mathbb{Z} / 2\mathbb{Z}$ with the subgroup ${0,8} \subset \mathbb{Z} / 16 \mathbb{Z}$, my choice of corresponding subgroup of automorphisms is still correct? Also, any advice on how to find a corresponding polynomial? – jonan Apr 02 '20 at 19:47