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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.

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    Why are you looking at _subgroups_? The Galois groups of subextensions of $\bar{K} \mid K$ are _quotients_ of the absolute Galois group.2012-05-07
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    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?2012-05-07
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    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.2012-05-07

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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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    thanks for the answer - CFT is actually where I'm headed, and I'm trying to make precise some of the "big picture" statements I've heard about it (for instance, that CFT studies abelian extensions). Could you provide a reference for or describe a proof of the statement that $\text{Gal}(K^\text{ab}/K) \cong \text{Gal}(\bar{K}/K)/[G,G]$ ?2012-05-07
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    A very pleasing answer.2012-05-07
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    @Lubin: In turn, a very pleasing comment. Thanks.2012-05-07
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    @mebassett: This is basically two things, the first is that the abelianization of a group is its maximal abelian quotient, and second is the fundamental theorem of Galois theory (so abelian subextensions correspond precisely to abelian quotients of the full Galois group).2012-05-07
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    I've been reading your answer to the above question posted last year and one thing that I don't quite understand is why the compositum of all the abelian subextensions of $K$ inside an algebraic closure $\bar{K}$ is also abelian. If you have two elements inside the compositum then would they necessarily belong to a same finite abelian subextension of $\bar{K}$? Many thanks for your help!2013-06-26
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    Yup. If $\alpha,\beta\in K^{ab}$, then the Galois closure of $K(\alpha,\beta)$ is an abelian extension of $K$.2013-06-26
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    This seems like a natural statement but I'm not sure how to prove it. Is there an easy way of seeing this? By the Galois closure I take it you mean the splitting field of the min pol of the primitive element generating $K(\alpha,\beta)$ over $K$? Thank you very much.2013-06-26
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    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!2013-06-26
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    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!2013-06-26
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    Maybe the issue is Galois closures? For $\alpha\in K^{ab}$, it's not even true that $K(\alpha)/K$ has to be Galois, 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.2013-06-26
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    Yes, my main query at the moment is with Galois closures although my original question was why $K^{ab}$ should be a number field. Could you please explain why the Galois closure of $K(\alpha)/K$ is a subfield of $K^{ab}/K$ if $\alpha \in K^{ab}$? Also does the Galois closure of a composite of two number fields (inside a fixed algebraic closure) equal the composite of their Galois closures? Thank you.2013-06-26
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    Sorry, my original question was why $K^{ab}$ should be an abelian extension of $K$.2013-06-26
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    Ah, yes. The compositum of two Galois closures is the Galois closure of the compositum. So since each individual Galois closure is abelian, so is their compositum (by arguments above).2013-06-26
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    Many thanks for your reply. That would certainly provide the final step in showing that the Galois closure of $K(\alpha,\beta)$ is an abelian extension of $K$ but could you explain why the Galois closure of $K(\alpha)/K$ is a subfield of $K^{ab}/K$ if $\alpha \in K^{ab}$? Thanks again.2013-06-27
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    If $K\subset L$, then $K^{gc}\subset L^{gc}$. Since $K\subset K^{ab}$, we have $K^{gc}\subset K^{ab.gc.}$. But since $K^{ab}$ is already Galois (being a compositum of Galois extensions), so $K^{ab.gc.}=K^{ab}$, and we're done.2013-06-27
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    I understand now. Many thanks again for your explanations- they've been very helpful.2013-06-28