Hi readers,

I have been studying Class Field theory this semester. Since I didn’t take the Algebraic number theory course offered in the previous semester, I am facing difficulty following the course. But this in no sense makes the course less interesting. Here in this post, I pen down a few thoughts on the subject, which are just my understanding of the subject. Note that WHAT FOLLOWS MAY BE INCORRECT, not to mention imprecise. But I will try and update this as my understanding increases over time.

In Class Field theory (CFT), the mission is to understand which is horrendously difficult, so they study , where is the maximal abelian extension of . By that, I mean such an (infinite) extension of so that its Galois group (which again is infinite) is abelian. Now for infinite Galois extensions the correspondence between subfields and subgroups is no longer true, so they do what they always do — change the definitions to make this correspondence work. One defines a topology on and now instead of subsets of this huge (uncountable) set, the correspondence is between closed sets and subgroups.

One also studies the field extension L/K where L and K are number fields instead of K/Q, as in a basic number theory course. Initially, I had difficulty assimilating the results and finding which results true for K/Q go through to L/K. For example, it every extension K/Q, at least one prime ramifies (in fact, all the primes dividing the discriminant). The units of Z are just 1 and -1 but there are lots more units in (given by Dirichlet in his Units theorem). As a consequence, it is possible that in an extension L/K of number fields, no prime of ramifies in . When the discriminant is a unit then no prime ramifies and the extension is called “unramified”.

To be continued..

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January 19, 2012 at 21:15

Dinesh valluri“By that, I mean such an (infinite) extension of \mathbb Q so that its Galois group (which again is infinite) is abelian”

….but there can be many such extensions for example $K = \mathbb Q(\mu_{infinity}$ which is the union of all cyclotomic extensions of $\mathbb Q$ is infinite extension with infinite Galois group but $\mathbb Q^{ab}$ contains $K$. So may be the definition of $\mathbb Q^{ab}$ should be revised?

January 19, 2012 at 21:32

dinesh1729According to wikipedia I may be wrong, it says that Q^{ab} is equal to K(as defined above) but still I think the definition should be revised or may be an answer to this question may clear things: Is Q^{ab} the only extension of Q whose galois group is abelian?

January 20, 2012 at 08:27

abhishekparabHi Dinesh,

You are right in that there are many abelian extensions, for example any rational square root adjoined to Q has Galois group of order 2, hence abelian. But when we speak of we mean a maximal such extension such that every finite abelian extension is hidden somewhere in this extension. Loosely speaking, it means taking the union of all abelian extensions of Q. For finite Galois extensions K and L of Q, we can form the composite extension KL whose Galois group is the product of the two Galois groups (when their intersection is Q). This process can be carried on using some kind of induction and the precise algebraic term is “taking direct limit”. So, is “the” maximal abelian extension.