This question is inspired by Emerton's question whether the ideal class of the different has a canonical square root.

Consider the diagram (of elements; the groups these lie in are the ideal class group of $L$ in the top row and that of $K$ in the bottom row) $$ \matrix{ ? & \to & [diff(L/K)] \cr \downarrow & & \downarrow \cr [St(L/K)] & \to & [disc(L/K)]} $$ where the horizontal maps are squaring and the vertical maps are taking norms from $L$ down to $K$. Here $diff(L/K)$ is the different of an extension of number fields, and $disc(L/K)$ its discriminant. The element $[St(L/K)]$ of the class group of the base field $K$ is the Steinitz class (see KConrad's comment to Emerton's question).

My question is whether there exists an element in the class group $Cl(L)$ that is at home in the left upper corner of this diagram. The most simple question would be whether the Steinitz class is always a norm; and if it is, whether it is the norm of a class whose square is the different class. The last question would be whether this element "?" is unique up to elements that lie in the intersection of the kernel of the norm and that of squaring.


Taking $L$ to be the Hilbert class field of $K$, such a construction would imply that the Steinitz class of $L/K$ is always trivial. Yet this is false - take $K = \mathbb{Q}(\sqrt{-15})$ for example.

(EDIT): If $L = K(\sqrt{\alpha})$ is a tamely ramified extension of $K$, then the Steinitz class is represented by an ideal $I$ such that $I^2 (\alpha) = \Delta_{L/K}$. In particular, if $L/K$ is unramified, so $(\alpha) = \mathfrak{n}^2$, then the Steinitz class is trivial if and only if $\mathfrak{n}$ is principal, i.e., if and only if we may take $\alpha$ to be a unit in $K$. Clearly this is not the case for $L/K$ above.

In fact, I just found a source that works out this example in explicit detail - see Theorems 2.2 and 3.1 of the following:


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    $\begingroup$ FJ strikes again! $\endgroup$ – Emerton Jul 30 '10 at 3:10
  • $\begingroup$ Is this really a counterexample? The Hilbert class field of K is generated by a square root of -3, and this extension seems to have a relative integral basis, which would imply that the Steinitz class is trivial. $\endgroup$ – Franz Lemmermeyer Jul 30 '10 at 9:17
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    $\begingroup$ Whoa, I wrote that thing and didn't even realize it gave a counterexample. $\endgroup$ – KConrad Jul 30 '10 at 15:07
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    $\begingroup$ Great. By now I also found a Ph.D. thesis by Rebecca Roy (tobias-lib.uni-tuebingen.de/volltexte/2003/864) that contains this example. It also studies the existence of "?", which is called a Steinitz root there. $\endgroup$ – Franz Lemmermeyer Jul 30 '10 at 16:08

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