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We all agree that the theory of class fields plays an eminent role in modern number theory.
Nevertheless, what was our main concern is that how to solve various Diophantine equations to which the clues do not appear clear in the theory of class fields. And my question is:

With respect to the beauty and perfection of the theory of class fields, can we use it to solve some mysterious Diophantine equations? Or can we apply it to other fields in Mathematics, such as Complex Analysis?

The last question arises as one of my friends used to tell me that there is a mathematician who developes the theory of complex analysis in terms of class fields in one of his books. Nonetheless, when I looked up to his books, I didn't find any hint to this. Moreover, when I asked him about this again, he didn't say much. This stimulates somehow my imagination such that I am wondering if there indeed is such a theory.
Any response is appreciated.

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You double posted; you should delete one of the two copies. –  Arturo Magidin Mar 12 '11 at 3:48
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@Arturo Magidin: I have no idea what happened, I am not good at computer, and the system just reminded me that there is an error, which is absolutely their fault, in any case, I will do so, and thanks. –  awllower Mar 12 '11 at 3:54
    
Combining with this, one can tell that I am confused and I would like to do some concrete exercises so that I can believe(or have some motivation for) the compelling theory of class fields. –  awllower Mar 12 '11 at 6:18

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Let me address the question as to whether class field theory can be used to solve some Diophantine equations. The answer is certainly yes. One historical example is given by Kummer's work on Fermat's Last Theorem; this relied on algebraic number theoretic results to do with cyclotomic fields, much of which can be reinterpreted as special cases of class field theory. (One place to see this discussed is the nice historical chapter on Kummer's work by Michael Rosen in the book Modular forms and Fermat's Last Theorem (Cornell, Silverman, Stevens eds.).)

At a more basic level, consider the question of solving the following three equations:

$$x^2 + y^2 = p$$ $$ x^2 + 5 y^2 = p$$ $$x^2 + 23 y^2 = p$$ where $p$ is a prime and $x$ and $y$ are integers.

Essentially (i.e. excluding some small primes, namely $p = 2$ in the first, $p = 2$ or $5$ in the second, and $p = 2$ or $23$ in the third) each of these questions can be rephrased as asking whether the prime $p$ splits into a product of principal ideals in the extension $K:= \mathbb Q(\sqrt{-D})$, where $D = 4,$ $20$, and $23$ respectively.

Now the question of whether $p$ splits is easily answered; it is just a question of whether the Jacobi symbol $\bigl( \frac{p}{D} \bigr)$ is equal to $1$ or not.

But the question of whether $p$ splits into a product of principal ideals is more subtle; it is a question of whether $p$ splits completely in the Hilbert class field $H$ of $K.$

Now in the case $D = 4$, we know that $K$ has class number one, and so $K = H$. Thus we can solve $x^2 + y^2 = p$ if and only if $p \equiv 1 \bmod 4$.

If $D = 20$, then $K$ has class number two, and in fact $H = K(\sqrt{-1})$ is the compositum of the field $\mathbb Q(\sqrt{-4})$ and $\mathbb Q(\sqrt{-20})$. Thus $p$ splits completely in $H$ if and only if $\bigl(\frac{p}{4}\bigr) = 1$ and $\bigl(\frac{p}{20}) = 1$, i.e. if and only if $\bigl(\frac{p}{4}\bigr) = 1$ and $\bigl(\frac{p}{5}) = 1$, i.e. if and only if $p \equiv 1 \bmod 4$ and $p \equiv \pm 1 \bmod 5$, i.e. if and only if $p \equiv 1 \text{ or } 9 \bmod 20$.

Finally, if $D = 23$ then $K$ has class number three, and so $H$ is not an abelian extension of $\mathbb Q$; rather it is an $S_3$-extension. The equivalence of class fields and abelian extensions says that $H$ is not a class field of $\mathbb Q$, and so there is no congruence condition on $p$ that determines whether or not $p$ splits completely in $H$. Thus there is no congruence condition on $p$ that determines whether or not we can solve $x^2 + 23y^2 = p$.

Instead, one has the following result from non-abelian class field theory (due, in this case, to Hecke, but best understood from a modern viewpoint as being a special case of Langlands's general program for non-abelian class field theory):

We can solve $x^2 + 23y^2 = p$ if and only if the coefficient of $q^p$ in the product $$q\prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}),$$ which is a priori equal to $-1,0,$ or $2$, is in fact equal to $2$. Computing the product, one finds e.g. that the first such prime is $p = 59 = 6^2 + 23\cdot 1^2$.

(This infinite product is a certain modular form, which is a particular kind of non-abelian analogue of a Jacobi symbol.)

Whether class field theory has applications to complex analysis, I don't know, but the appearance of modular forms at the end of the above discussion shows that complex analysis has applications to class field theory.

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@Matt E: It's an excellent explanation, and it does clarify some ambiguous notion of me. However, I must say, that I didn't mean to cause anyone's problem; rather, I kind of want to make an analogue to the usage, or idiom, that one sometimes says, instead of algebraic number theory, the theory of algebraic number fields. –  awllower Mar 12 '11 at 9:58
    
@awllower: Dear Awllower, You didn't cause any problems, and I'm sorry for being rude regarding English usage. Best wishes, –  Matt E Mar 12 '11 at 15:53
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@Zev: Also, as a general remark: In a modern treatment of class field theory, one tends to focus on the idea of abelian extensions of number fields, and the fact that class field theory describes such extensions. This sits well with a typical undergraduate background in algebra and (especially) Galois theory, but is only part of the picture. In order to understand better what class field theory is about, it is worth making a real effort to understand the idea of class fields (as extensions in which splitting behaviour is determined by congruence conditions), so that ... –  Matt E Mar 13 '11 at 5:43
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... class field theory becomes the statement that class fields of any conductor exist, and are precisely the abelian extensions. Thus an arithmetic notion (class fields) is (highly non-obviously) equivalent to an algebraic one (abelianness of a field extension). Of course, this is what class field theory is all about, and so it is described in some fashion in every treatment of the subject. But it may not always be brought to the fore, especially in modern treatments where the notion of abelian extension, giving rise to a well-defined Artin map, comes first, rather than the notion of ... –  Matt E Mar 13 '11 at 5:47
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... class field. Regards, –  Matt E Mar 13 '11 at 5:48

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