What is the Hilbert class field of a cyclotomic field? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:17:49Z http://mathoverflow.net/feeds/question/968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field What is the Hilbert class field of a cyclotomic field? Ben Webster 2009-10-18T01:04:09Z 2009-10-18T20:49:38Z <p>In the answers to <a href="http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers" rel="nofollow">Qiaochu's post</a> on defining representations of finite groups over the algebraic integers, it came out that which fields a representation of a finite group is defined over might depend on whether you require your representation to be free or just projective. If your representation V is defined over a number field K, it contains a projective submodule V' for the integers O of K such that V'\otimes_O K=V, but it's not at all clear if this can be chosen to be free. </p> <p>Luckily, there is a very nice theory of algebraic number theory that says that any projective module over the ring of integers of a number field becomes free when you extend scalars to the Hilbert class field. So, since all representations of finite groups are defined over cyclotomic fields (in fact, you just need the roots of unity for the orders of elements in G), every representation has as an integral basis in the Hilbert class field of a cyclotomic field. Which is.....?</p> <p>Edit: while the class numbers are interesting that's not the question I asked. I want to actually what the Hilbert class field is, or something about it. For example, is it cyclotomic (seems unlikely, but cyclotomic fields are nice...)?</p> http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field/972#972 Answer by David Zureick-Brown for What is the Hilbert class field of a cyclotomic field? David Zureick-Brown 2009-10-18T01:29:58Z 2009-10-18T01:29:58Z <p>They can get arbitrarily large. You can write down formulas in certain cases; this a main theorem of Iwasawa theory. See also the notion of a regular prime.</p> http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field/976#976 Answer by Andreas Holmstrom for What is the Hilbert class field of a cyclotomic field? Andreas Holmstrom 2009-10-18T01:49:18Z 2009-10-18T02:39:53Z <p>It follows from class field theory that for any number field K, there is an isomorphism between the class group of K and the Galois group Gal(H/K), where H is the Hilbert class field of K. In particular, the degree of H over K is the class number of K. As David points out, class numbers of cyclotomic fields are complicated, and closely related to classical Iwasawa theory, see for example the book of Washington on cyclotomic fields. </p> <p>Example: If you take K = Q(zeta) where zeta is a p-th root of unity, p an odd prime, then the class number of K tends to grow with p. In particular, the class number is one iff p &lt; 20. </p> <p>One the other hand, if you fix p, and study the fields K_n obtained by adjoining a p^n-th root of unity to Q, then I believe that the exponent of p in the class number is independent of n (at least for n large enough), and the exponent of any other prime is bounded as n goes to infinity. Maybe one can say something more precise about these exponents.</p> http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field/999#999 Answer by Lavender Honey for What is the Hilbert class field of a cyclotomic field? Lavender Honey 2009-10-18T04:32:50Z 2009-10-18T04:32:50Z <p>Giving an "explicit" description of the Hilbert class field of a number field K (or, more generally, all abelian extensions of K) is Hilbert's 12th problem, and has only been solved for Q and for imaginary quadratic fields. The Hilbert class field H of Q(zeta) will only be contained in a cyclotomic field if H = Q(zeta) itself --- since one can explicitly compute that any abelian extension of Q properly containing Q(zeta) is ramified at some place.</p> http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field/1012#1012 Answer by Joel Dodge for What is the Hilbert class field of a cyclotomic field? Joel Dodge 2009-10-18T07:20:13Z 2009-10-18T07:34:45Z <p>". . .if you fix p, and study the fields K_n obtained by adjoining a p^n-th root of unity to Q, then I believe that the exponent of p in the class number is independent of n (at least for n large enough). . ."</p> <p>The correct (but still vague) statement here would be, not that the p part of the class numbers are independent of n for large n, but that the growth of the class numbers can be described very explicitly in terms of n. Roughly speaking: the p-part of the class number of K_n has exponent mp^n+ln+v for some integers m, l, v. </p> http://mathoverflow.net/questions/968/what-is-the-hilbert-class-field-of-a-cyclotomic-field/1070#1070 Answer by Jonah Sinick for What is the Hilbert class field of a cyclotomic field? Jonah Sinick 2009-10-18T20:49:38Z 2009-10-18T20:49:38Z <p>Since the Hilbert class field is defined to be the maximal <em>unramified</em> extension of a number field, I think it should be easy to see that the Hilbert class field of a given cyclotomic field is not cyclotomic unless the class number of the base field is 1/</p> <p>Ken Ribet has a paper on a related matter</p> <p>Ribet, Kenneth A. A modular construction of unramified p-extensions of Q(µ_p). Invent. Math. 34 (1976), no. 3, 151--162. </p> <p><a href="http://math.berkeley.edu/~ribet/Articles/invent_34.pdf" rel="nofollow">http://math.berkeley.edu/~ribet/Articles/invent_34.pdf</a></p>