How does one know that a number field $K$ has a maximal abelian extension (unique up to isomorphism) $K^{\text{ab}}$?

I've read proofs involving Zorn's lemma that it has an algebraic closure (And that algebraic closures are unique up to isomorphism.) $\bar{K}$ All these proofs involved ideals of the polynomial ring in variables $x_f$, $f$ an irreducible monic polynomial in $K[x]$, but I don't see any obvious way of "restricting" this proof to abelian extensions.

I tried proving that such an extension exists using Zorn's lemma: Let $\Sigma$ be the set of all abelian subgroups of $\text{Gal}(\bar{K}/K)$ partially ordered by inclusion. Any chain of subgroups $(G_\alpha)$ has an upper bound, namely, $\bigcup_\alpha G_\alpha$ (which is a [sub]group as each $G_\alpha$ is contained in another), so by Zorn's lemma $\Sigma$ has a maximal element. But I don't have that this element is unique. (and I don't think I proved that $\bigcup_\alpha G_\alpha$ is abelian, either).

Additionally, how does $\text{Gal}(K^\text{ab}/K)$ relate to $\text{Gal}(\bar{K}/K)$ ? My incomplete attempt at a Zorn's lemme proof doesn't tell me what the maximal abelian galois group should be, and I don't know many ways of finding abelian subgroups.

  • $\begingroup$ Why are you looking at subgroups? The Galois groups of subextensions of $\bar{K} \mid K$ are quotients of the absolute Galois group. $\endgroup$
    – Zhen Lin
    May 7 '12 at 15:16
  • $\begingroup$ Perhaps because I'm naive. :) But is not looking at quotients of $\text{Gal}(\bar{K}/K)$ equiv to looking at the normal subgroups (except that the inclusion is reversed) ? or am I missing something subtle about galois theory? $\endgroup$
    – mebassett
    May 7 '12 at 15:25
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    $\begingroup$ An abelian extension of the base field doesn't correspond to an abelian subgroup but to an abelian quotient. So the maximal abelian extension has Galois group the abelianization of the absolute Galois group. $\endgroup$ May 7 '12 at 15:37
  • $\begingroup$ @QiaochuYuan, what do you mean by abelian quotient ? Does it mean a quotient of an abelian group which may not be abelian ? $\endgroup$
    – Why
    Aug 8 at 6:55
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    $\begingroup$ @Why let G be a group and H be a normal subgroup. You have the quotient group G/H. If that quotient is abelian we might call it an abelian quotient. G need not be abelian. Indeed, if G is non-abelian, you can let H = [G,G] (the commutator subgroup) and then G/H is called the abelianization of G. $\endgroup$
    – mebassett
    Aug 10 at 14:45

Existence: It's not hard to check that a compositum of abelian extensions is again abelian (the Galois group of the compositum embeds into the product of the individual Galois groups) and the maximal abelian extension of $K$ is precisely the compositum of all such extensions. Once you've constructed an algebraic closure, you don't have to worry about working directly with minimal polynomials.

Relation to $\operatorname{Gal}(\overline{K}/K)$: As mentioned in the comments, the Galois group of the maximal abelian extension is just the abelianization of the absolute Galois group. Note that this abelianization process can be trivial (though not for number fields) -- if you start with a finite field, or $\mathbb{R}$, its algebraic closure is already an abelian extension.

What it looks like: Remarkably, this is a largely wide open question, and I'll just briefly reference you to the whole branch of number theory known as class field theory. When $K=\mathbb{Q}$, the answer is completely understood (but fairly non-trivial): The maximal abelian extension is the field obtained by adjoining all roots of unity to $\mathbb{Q}$, i.e., the splitting field of the set of polynomials $x^n-1$ for all $n\geq 1$. The Galois group is precisely $\prod \mathbb{Z}_p^\times$, where the product ranges over all primes $p$. (Note this is an uncountable group). There's also an explicit version of such a statement in the case that $K$ is quadratic imaginary, where the maximal extension is obtained by adjoining special values of functions defined on elliptic curves. Beyond those two cases, the state of the art ranges from fairly explicit conjectures (e.g., Stark conjectures for totally real fields, in particular real quadratic fields) to completely unknown. That said, there's lots of neat stuff known about these fields and their Galois groups which falls shy of an explicit construction, but I suspect they lie beyond the scope of the answer you were looking for.

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    $\begingroup$ A very pleasing answer. $\endgroup$
    – Lubin
    May 7 '12 at 18:43
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    $\begingroup$ Yup. If $\alpha,\beta\in K^{ab}$, then the Galois closure of $K(\alpha,\beta)$ is an abelian extension of $K$. $\endgroup$ Jun 26 '13 at 16:24
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    $\begingroup$ The Galois group of a compositum of two number fields is a subgroup of the direct product of those two fields (in fact, it's the subgroup that fixes the intersection). The product of abelian groups is abelian, and subgroups of abelian groups are abelian. That's it! $\endgroup$ Jun 26 '13 at 17:44
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    $\begingroup$ So is it true that if $\alpha \in K^{ab}$ then $K(\alpha)/K$ is an abelian extension? If so, why? If this is true then since $K(\alpha,\beta)$ is equal to $K(\alpha)\cdot K(\beta)$ the former field would be an abelian extension of $K$ by your comments? I'm afraid I still can't see why the Galois closure of $K(\alpha,\beta)$ is an abelian extension of $K$. Many thanks! $\endgroup$
    – Josh F.
    Jun 26 '13 at 18:02
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    $\begingroup$ Maybe the issue is Galois closures? For $\alpha\in K^{ab}$, it's not even true that $K(\alpha)/K$ has to be <i>Galois</i>, let alone abelian. There's also a bit of confusion about the order of the construction. If $\alpha\in K^{ab}$, then the Galois closure of $K(\alpha)/K$ is a subfield of $K^{ab}/K$, so its Galois group is a quotient of an abelian group, so is abelian. Of course, you have to believe $K^{ab}$ already exists to make this argument, which is what I thought was what your question was really about. $\endgroup$ Jun 26 '13 at 18:18

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