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I usually consider a cyclic extension $K$ of degree an odd prime $p$ over the rational field $\mathbf{Q}$.

In this case, there is a well-known result that "every ambiguous class in the class group $\operatorname{Cl}_K$, which is a class fixed by the Galois group of $\operatorname{Gal}(K/\mathbf{Q})$, becomes trivial in the genus field of $K/\mathbf{Q}$". Two references are Terada and Furuya.

Let $E$ be the maximal unramified extension of $K$ whose Galois group over $K$ is isomorphic to an elementary abelian $p$-group. In my opinion, the analogue of the principal ideal theorem should hold for $E/K$. More precisely, I guess the following statement is true:

Every ideal $I$ of $K$ such that $I^p$ is principal (in $K$) becomes principal in $E$.

I tried to modify the method of Terada or Furuya, and failed. (Because, in some sense, a symmetric of the form of $I^p$ is less than $I^ {(1-\sigma)}$ where $\sigma$ is a generator of $\operatorname{Gal}(K/\mathbf{Q})$ and there is an explicit expression about the genus field but not $E$.)

If you know about a method for this type of problems, related good references, counter-examples, or have some comments, please let me know.

Sorry for my poor English. Thank you for your attention.

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If "can be" means "is", then there are counterexamples in Furtwängler's article from 1916 (Monatsh. f. Math. 27). –  Franz Lemmermeyer Nov 10 '13 at 12:29
    
@FranzLemmermeyer: I have downloaded the paper and read it, but found no counterexample. It seems to me that the paper only deals with the (by then not yet proved) classical principal ideal theorem in the case $Cl_k=(\mathbb{Z}/\ell)^2$, proving it. The various examples treat capitulation in the intermediate fields of degree $\ell$, but these do not fall into the OP setting. –  Filippo Alberto Edoardo Nov 10 '13 at 15:44
    
@Filipo: sorry, my mistake. So the smallest counterexample would be a (2,4)-extension such that only an ideal of order 4 capitulates in the genus field. It can't be that hard to find. –  Franz Lemmermeyer Nov 10 '13 at 20:41
    
@Filipo: Thank you!! –  kjs Nov 11 '13 at 2:21
    
@FranzLemmermeyer: Thank you for your comment. I was originally interested in the case Gal($K$/$Q$) = (p) where p is odd prime and didn't consider the cases in the paper you mentioned. I'm a little bit confused about the definition, though. From your survey paper "class field towers(2010)", it seems (2,4)-extension means a field $K$ with Gal($K$/$Q$) = (2,4). Am I right? If so, can you suggest a nice example of such field that may be easy to compute? –  kjs Nov 11 '13 at 2:44

1 Answer 1

I have been trying to prove the result for a couple of days, without success, so I post what I got in the meanwhile. Let me suppose throughout that $\operatorname{Gal}(E/K)\cong(\mathbb{Z}/p)^2$ (the case $E/K$ cyclic is solved by Hilbert 94).

As Franz Lemmermeyer noticed, the answer is clear when $E$ is the Hilbert class field of $K$ by the classical principal ideal theorem, so let me suppose that we are in the first non-trivial instance, namely $$ \mathrm{cl}_K\cong\mathbb{Z}/(p)\times\mathbb{Z}/(p^2) $$ (I am implicitely killing everything which is prime-to-$p$, since the problem is stable under restriction to one $p$-component at a time).

Lemma: If the Hilbert class field of $E$ and of $K$ coincide, namely if $\mathrm{cl}_E\cong\mathbb{Z}/(p)$, then every ideal class of $\mathrm{cl}_K$ is principal in $E$.

Proof Call $H=H_K=H_E$ the Hilbert class field of $K$ (or of $E$, by hypothesis): it has degree $p$ over $E$ and degree $p^3$ over $K$. Let $c=[\mathfrak{p}]$ be a class of order $p^2$ in $\mathrm{cl}_K$. Then its inertia degree in $H/K$ is $p^3$ but it cannot stay inert in $E/K$ because $\mathrm{Gal}(E/K)$ is not cyclic, hence it is not isomorphic to the Galois group of an extension of the finite field $\mathcal{O}_K/\mathfrak{p}$, as it would be the case if $\mathfrak{p}$ were inert. It follows that $\mathfrak{p}\mathcal{O}=\mathfrak{P}_1\cdots\mathfrak{P}_p$ and the primes $\mathfrak{P}_i$ are all conjugate under $\mathrm{Gal}(E/K)$. Hence, they all belong to the same class in $\mathrm{cl}_E\cong\mathbb{Z}/(p)$ (and an easy argument shows this class is non-trivial, but it is not needed here), and $[\mathfrak{p}\mathcal{O}_E]=[\mathfrak{P}_1]^p=[0]$. Thus the initial class $c$ capitulates, and so does every class of order $p^2$ since $c$ was arbitrary. As $\mathrm{cl}_K$ is generated by classes of order $p^2$, everything becomes principal in $E$, q.e.d

Of course, the above proof makes crucial use of the very unnatural assumption that $\mathrm{cl}_E=\mathbb{Z}/(p)$ and I quite agree with Franz Lemmermeyer that it should be possible to come up with an exemple where one element of order $p$ in $\mathrm{cl}_K$ do not capitulate, although I am been unable to do so. May be, the only thing you can save of my answer (but which is already contained in Furtwängler's paper quoted in the comments) is that the structure of the class group of $E$ plays a crucial role for your problem. On the other hand, I doubt that $K/\mathbf{Q}$ being cyclic of order $p$ tells you much.

ADDENDUM $7^\text{th}$ February $2014$

A paper with an explicit counterexample appears today on the arXiv, here and was published in 2008 in Acta Arithmetica.

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The result that if the Hilbert class field tower terminates, then all ideals that are allowed to capitulate by group theoretical reasons do capitulate is classical. For cyclic extensions of odd prime degree the result might actually hold - if you are lloking for counterexamples, ask Daniel Mayer algebra.at/index_e.htm if he knows one. –  Franz Lemmermeyer Nov 16 '13 at 17:51
    
For finding a counterexample in the case of (2,4)-extensions, a first step would be finding a group theoretical counterexample be letting GAP search for a group with a corresponding kernel of the associated transfer maps. Armed with the smallest such group a corresponding extension can perhaps be constructed. The other option is a brute force search. –  Franz Lemmermeyer Nov 16 '13 at 17:52

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