most recent 30 from http://mathoverflow.net 2013-05-24T14:46:07Z http://mathoverflow.net/feeds/question/74373 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74373/lower-bound-on-the-class-group-of-the-p-hilbert-class-field-of-an-imaginary-quadr Tobias Bembom 2011-09-02T16:18:40Z 2011-09-02T16:44:48Z

<p>Let K be an imaginary quadratic field, A(K) its p-class group, and H(K) its p-Hilbert class field. If rk(A(K))=2, a result due to Arrigoni tells us that p^3 divides the order of the class group of H(K). Are there any explicit non-trivial lower bounds in the case that rk(A(K))>2 ?</p>

http://mathoverflow.net/questions/74373/lower-bound-on-the-class-group-of-the-p-hilbert-class-field-of-an-imaginary-quadr/74377#74377 Cam McLeman 2011-09-02T16:44:48Z 2011-09-02T16:44:48Z

<p>If $G$ is the Galois group of the $p$-class field tower over $K$, then $A(H(K))=G'/G''$ is a quotient of $G_2/G_4$, where $G_i$ denotes the lower central series. By Arrigoni's calculation that $G_2/G_4$ has $p$-rank exactly $\frac{d(d-1)(2d+5)}{6}$, this serves as a lower bound for the $p$-rank of $A(H(K))$. When $d=2$, you get the bound of $3$ you mention in the question. Note that the calculation is actually much more precise: The size of $A(H(K))$ depends not only on the rank, but on the orders of the generators of the $p$-class group. This will give you a better bound for the class number than simply raising $p$ to the rank bound given above.</p>