# Module of Kahler differentials of rings of integers of number fields

How does one prove that if $L/K$ is an extension of number fields with rings of integers $B/A$, then the module of Kahler differentials $\Omega^1_{B/A}$ can be generated by one element as a $B$-module? When one proves that the annihilator of $\Omega^1_{B/A}$ is the different of the extension, one localizes and completes since both the different and the module of Kahler differentials are compatible with localization and completion. In the local complete case one has $B = A[b]$ for some $b\in B$ and both the claim about the different and my question are easy. What I don't see is how to pick a $b\in B$ before localization and completion which would "work at all finite places" (assuming this approach could be used to solve my question, to begin with)?

-

## 1 Answer

You don't have to pick a $b\in B$ that works at all finite places. Indeed, such a $b$ might not exist. (I think there's an MO question about that.) But by the Chinese remainder theorem, a module over $B$ is cyclic if and only if it is so locally. So you can localize $B$, and then it's enough to check on the completion, where as you point out, it is true.

-
Thanks! For those who might read this in the future: I found en.wikipedia.org/wiki/… useful to understand the "by the Chinese remainder theorem" part. – Kestutis Cesnavicius Jul 12 '11 at 13:13