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I'm trying to prove that every finite abelian group is the Galois group of of some finite extension of the rationals. I think I'm almost there.

Given a finite abelian group $G$, I have constructed field extensions whose Galois groups are the cyclic groups occurring in the direct product of $G$. How do I show that the compositum of these fields has Galois group $G$.

Cheers

Arturo Magidin
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daisy
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    You can't show that because it isn't necessarily true. For example, the compositum of a quadratic extension $K$ with itself still has Galois group $C_2$ rather than $C_2 \times C_2$. – Qiaochu Yuan Apr 13 '12 at 16:49
  • Hint. If $A$ is an abelian group, and $H$ is a subgroup of $A$, then there is a subgroup $K$ of $A$ such that $A/K\cong H$. – Arturo Magidin Apr 13 '12 at 17:19
  • I date myself with this one, but you might find Theorem 5 and its first corollary of interest ( in the 1965 edition of Lang's Algebra). – Chris Leary Apr 13 '12 at 17:26
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    Why don't you use the fact that $Gal( \mathbb{Q}(\zeta_n) / \mathbb{Q})$ is isomorphic to the group of units of $\mathbb{Z}/n\mathbb{Z}$ ? find an appropriate $n$ such that the finite abelian group $G$ is a quotient of $\mathbb{Z}/n\mathbb{Z}$ and use the galois correspondence to prove the existence of such field. – Dinesh Apr 13 '12 at 17:55

4 Answers4

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The simplest way I know to do this is what Dinesh suggests in the comment, together with my hint above.

  1. Show that if $n\geq 1$, then the $n$th cyclotomic field $\mathbb{Q}(\zeta_n)$ (where $\zeta_n$ is a primitive $n$-th root of unity) has Galois group isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$. The structure of $(\mathbb{Z}/n\mathbb{Z})^*$ is well-understood in terms of the prime factorization of $n$.

  2. Given a finite abelian group $G$, find an $n$ such that $G$ is a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$.

  3. Prove that if $A$ is an abelian group, and $H$ is a subgroup of $A$, then there exists a subgroup $K$ of $A$ such that $A/K\cong H$.

  4. Use the Fundamental Theorem of Galois Theory and the points above to obtain the desired result.

Arturo Magidin
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    @ Arturo: For step 2 my group theory knowledge is lacking. By what result does a finite abelian group occur as a subgroup of $(\Bbb{Z}/n\Bbb{Z})$∗ for some n? Other than that I can complete the argument. ... The steps I was following before can be found here: http://www.math.tifr.res.in/~eghate/kw.pdf (See the discussion proceeding theorem 2). I would like to understand this as I see it as a nice example to some ramification theory that I have been studying recently. I am only unsure on the final bit, "Then some Galois theory shows Gal(K/Q) = G, and we are done" Thanks for any help – daisy Apr 14 '12 at 10:51
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    @daisy: The fundamental theorem of finitely generated abelian groups tells you that $G$ can be written as a direct product of cyclic groups of certain orders, $G\cong C_{n_1}\times\cdots\times C_{n_k}$ with $n_1|n_2|\cdots|n_k$. All you need to do is find primes $p_i$ such that $n_i|p_i-1$, and there are infinitely many such primes for each $i$ by Dirichlet's Theorem of primes in arithmetic progressions. Then $n=p_1\cdots p_k$ will do. – Arturo Magidin Apr 14 '12 at 19:40
  • @daisy: The "some Galois theory" is the inclusion reversing correspondence between subgroups and subfields. The linear independence translates into the fact that the subgroups intersect trivially and generate the entire group $\mathrm{Gal}(K/Q)$. This in turn tells you that $\mathrm{Gal}(K/Q)$ is isomorphic to the product of the corresponding subgroups. – Arturo Magidin Apr 14 '12 at 22:55
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    In fact, one does not need step 3, as one may take $G$ to be a quotient group from the beginning: $$G \simeq \prod_{i=1}^k \mathbb Z/n_i\mathbb Z \twoheadleftarrow \prod_{i=1}^k \mathbb Z/(p_i-1)\mathbb Z \simeq \left( \mathbb Z/p_1\ldots p_k\mathbb Z \right)^*.$$ – evgeny Nov 13 '16 at 10:54
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All that u have to use from the group structure of (Z/nZ)* is that it is direct product of cyclic groups. To find the appropriate $n$ use the dirichlet's theorem from which we can say there are infinitely many primes which are 1(mod n). And the thing about 'final bit' it is just the fundamental theorem of Galois theory. When you see that the $G$ can be seen as a quotient of $(Z/nZ)$* then according to the fundamental theorem there exists an intermediate field say $K$ between $\mathbb{Q}(\zeta_n)$ and $\mathbb{Q}$ such that $Gal(K/Q)$ is $G$. Now the only task remaining for you is showing that $G$ can be seen as a quotient of $(Z/nZ)$* using dirichlet's theorem.

Good luck.

Dinesh
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@Daisy, your question is part of something more general called the Inverse Galois Problem, which goes back all the way to David Hilbert and Emmy Noether: which finite groups can be realized as Galois group over a general field?. See for example the discussion https://mathoverflow.net/questions/6743/the-inverse-galois-problem-what-is-it-good-for. There exists a vast literature on the subject.

Community
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Nicky Hekster
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This is an assignment I have to do for my Galois Theory course. I reproduce the version I wrote a couple of days ago in the hope it may help other people.

Let $G$ be a finite abelian group. Show that there exists a Galois extension $K/\mathbb{Q}$ with $Gal(K/\mathbb{Q}) \approx G$.

Let $G \approx \mathbb{Z}_{n_1} \times \ldots \times \mathbb{Z}_{n_k}$ be any abelian group. In this case, by Dirichlet's Theorem, there exist distinct primes $p_i$ such that $p_i \equiv 1 \mod(n_i)$. If $n = p_1 \ldots p_k$, the Chinese Remainder Theorem tells us that: $$\mathbb{Z}_n \approx \mathbb{Z}_{p_1} \times \ldots \times \mathbb{Z}_{p_n}$$

Considering the restriction of the isomorphism from the Chinese Remainder Theorem to non-zero elements, we obtain that this ring isomorphism can be viewed as an isomorphism in the multiplicative group, i.e.: $$(\mathbb{Z}_n)^* \approx (\mathbb{Z}_{p_1})^* \times \ldots \times (\mathbb{Z}_{p_n})^*$$

Every cyclic group is isomorphic to $\mathbb{Z}$ or $\mathbb{Z}_k$ for some $k$. $(\mathbb{Z}_p)^*$ corresponds to a cyclic group of order $p-1$, i.e., in additive notation, we have: $$(\mathbb{Z}_n)^* \approx \mathbb{Z}_{p_1-1} \times \ldots \times \mathbb{Z}_{p_n-1} = \mathbb{Z}_{j_1n_1} \times \ldots \times \mathbb{Z}_{j_kn_k}$$

By Corollary 7.8 of Morandi's Galois Theory book, if $w$ is a primitive $n$-th root of unity, i.e., $w = e^{\frac{2\pi i}{n}}$, then $\mathbb{Q}(w) = \mathbb{Q}_w$ satisfies the following: $$Gal(\mathbb{Q}_w/\mathbb{Q}) \approx (\mathbb{Z}_n)^* \approx \mathbb{Z}_{j_1n_1} \times \ldots \times \mathbb{Z}_{j_k n_k}$$

We will verify that: $\mathbb{Z}_{j_1} \times \ldots \times \mathbb{Z}_{j_k} \approx \hat{G} \triangleleft Gal(\mathbb{Q}_w/\mathbb{Q})$. For this, let's recall the following general result about groups: if $H \approx \mathbb{Z}_u$ is cyclic of order $u$, then there exists a subgroup for each divisor of $u$. Let $d$ be such a divisor, and let $L \approx \mathbb{Z}_d$ be the associated subgroup. There are $u/d$ cosets by Lagrange, so $|H/L| = u/d$. Clearly, $H/L$ is generated by $xL$, where $x$ generates $H$. Thus, $H/L$ is cyclic and isomorphic to $\mathbb{Z}_{u/d}$.

From what we discussed, since each $j_k | j_kn_k$, we need $\mathbb{Z}_{j_k} \approx {G}_k \triangleleft \mathbb{Z}_{j_kn_k}$.

Thus, we conclude that $\mathbb{Z}_{j_1} \times \ldots \times \mathbb{Z}_{j_k} \approx {G}_1 \times \ldots \times {G}_k \triangleleft \mathbb{Z}_{j_1n_1} \times \ldots \mathbb{Z}_{j_kn_k} \approx Gal(\mathbb{Q}_w/\mathbb{Q})$, and by normality, we use the Fundamental Theorem of Galois Theory to show that $K = \mathcal{F}(\hat{G})$ is Galois over $\mathbb{Q}$, and also: $$Gal(K/\mathbb{Q}) \approx Gal(\mathbb{Q}_w/\mathbb{Q})/\hat{G} \approx \mathbb{Z}_{j_1n_1} \times \ldots \times \mathbb{Z}_{j_kn_k}/{G}_1 \times \ldots {G}_k $$

Because each $G_i$ is normal in $\mathbb{Z}_{j_in_i}$ we may simplify this further taking the quotient inside of the cross products!

$$Gal(K/\mathbb{Q})\approx\mathbb{Z}_{j_1n_1}/G_1 \times \ldots \times \mathbb{Z}_{j_kn_k}/G_k $$

Finally, because $\mathbb{Z}_{j_in_i}/G_i$ is cyclic and we know how many cosets there are by Lagrange ($j_i n_i/j_i$), we need:

$$Gal(K/\mathbb{Q})\approx\mathbb{Z}_{n_1} \times \ldots \times \mathbb{Z}_{n_k} \approx G$$

Kadmos
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