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Let $K$ be an algebraic number field, i.e. a finite extension of $\mathbb{Q}$. Let $G$ be a finite abelian group. Does there exist a Galois extension of $K$ whose Galois group is isomorphic to $G$? I can prove this if $K$ is $\mathbb{Q}$ by using a special case of Dirichlet's prime number theorem, i.e. there are infinitely many prime numbers $p$ such that $p \equiv 1$ (mod $n$) for a given integer $n \ge 1$. So if $|G|$ is relatively prime to $[K: \mathbb{Q}]$, there exists such an extension, but I have no idea otherwise.

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Caveat: an incomplete draft

Let $m=[K:\Bbb{Q}]$. By the structure theorem of finite abelian groups there exists integers $d_1\mid d_2\mid \cdots\mid d_k$ such that $$ G\cong C_{d_1}\times C_{d_2}\times \cdots\times C_{d_k}. $$ Choose distinct primes $p_1,p_2,\ldots,p_k$ such that $p_i\equiv1\pmod{md_i}$. Let $\zeta_i$ be a complex primitive root of unity of order $p_i$. Consider the extension $K(\zeta_i)/K$. As a splitting field of a cyclotomic polynomial $K(\zeta_i)$ is Galois over $K$. Because any $K$-automorphism of $K(\zeta_i)$ is determined by the image of $\zeta_i$, the group $Gal(K(\zeta_i)/K)$ is isomorphic to a subgroup $G_i$ of $Gal(\Bbb{Q}(\zeta_i)/\Bbb{Q})\cong C_{p_i-1}.$ The degree of the extension $$ [K(\zeta_i):K]=\frac{[K(\zeta_i):\Bbb{Q}]}{[K:\Bbb{Q}]}= \frac{[K(\zeta_i):\Bbb{Q}(\zeta_i)]\cdot(p_i-1)}{m} $$ is a multiple of $d_i$. Therefore $d_i\mid |G_i|$. This implies that there is an intermediate field $K_i, K\subseteq K_i\subseteq K(\zeta_i)$ such that $K_i/K$ is cyclic of order $d_i$.

If we can show that the extensions $K_i/K$ are linearly disjoint, then we reach the conclusion that their compositum $\tilde{K}=K_1K_2\cdots K_k$ is a Galois extension with $Gal(\tilde{K}/K)\cong G$. Initially I felt that this follows immediately from the fact that the extensions $\Bbb{Q}(\zeta_i)/\Bbb{Q}, i=1,2,\ldots,$ are linearly disjoint over the rationals. Unfortunately I cannot justify this at the moment. Leaving this as incomplete for now.

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  • $\begingroup$ May be simpler to build it without a compositum using a subfield of $K(\zeta)$ with $\zeta$ of order $p_1p_2\cdots p_k$? $\endgroup$ – Jyrki Lahtonen Nov 25 '15 at 7:30
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Yes, it exists. First adjoin a root of unity, of sufficiently large degree, to get a cyclotomic extension. The Galois group will be cyclic. All intermediate fields will be Galois, and now try to arrange this intermediate field to be isomorphic to the group that you want.

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  • $\begingroup$ I don't understand. Would you elaborate on it? $\endgroup$ – Makoto Kato May 28 '15 at 0:18

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