Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p$ be an odd prime. Let $K_{\infty}$ be the field extension of $\mathbb{Q}$ generated by all $p^n$-th roots of unity. Let $M/K_{\infty}$ be the maximal abelian $p$-extension of $K_{\infty}$ unramified outside $p$. Denote its Galois group by $\mathcal{X}$. Then the Iwasawa main conjecture concerns the structure of $\mathcal{X}^{+}$ (here, $\pm$ means the $\pm1$-eigenspace of complex conjugation). My question is whether one can say anything about the structure of $\mathcal{X}^-$?

I don't find the answer in the basic textbook (e.g. Lang's Cyclotomic fields & Washington's Introduction to cyclotomic fields). I wonder whether $\mathcal{X}^-$ is a free $\mathbb{Z}p[[\Gamma]]$-module where $\Gamma$ is the Glois group of $K_{\infty}/\mathbb{Q}(\zeta_{p})$?

Can somebody tell me the answer? Thank you very much.

share|improve this question

2 Answers 2

up vote 7 down vote accepted

By Theorem 13.31 in L. Washington Introduction to Cyclotomic Fields, Second Edition, GTM 83 we know that $\mathcal{X}\sim \Lambda^{r_2}\oplus(\Lambda-$torsion), where $r_2$ is the number of complex embeddings of $\mathbb{Q}(\zeta_p)$, so $r_2=(p-1)/2$, and $\sim$ means "pseudo-isomorphism". In particular, since we know that the plus part is torsion, one deduces that the minus part of $\mathcal{X}$ is huge (finitely generated but not torsion over $\Lambda$).

But observe that $\mathcal{X}^+$ "is dual" to $X^-$ where I denote by $X=\mathrm{Gal}(L/K_{\infty})$ the Galois group of the maximal abelian $p$-extension of $K_{\infty}$ which is unramified everywhere. As for my "duality" statement, we have (always quoting Washington)

  • Proposition 13.32 $e_j\mathcal{X}(-1)\cong \mathrm{Hom}_{\mathbb{Z}_p}(e_iA_\infty,\mathbb{Q}_p/\mathbb{Z}_p)$ where $i$ is odd and $i+j\equiv 1\pmod{p-1}$ (you can always assume both $0\leq i,j\leq p-1$ so $j=p-i$ is even and the above is a statement about a piece of $\mathcal{X}^+$ and a piece of $X^-)$; and $A_\infty$ is the inductive limit of $p$-Sylow of class groups along the cyclotomic extension.

  • Proposition 15.34 $\tilde{X}\sim (A_\infty,\mathbb{Q}_p/\mathbb{Z}_p)$ where $\tilde{X}$ is the old $X$ with twisted $\Gamma$-action (see p. 359).

Combining the two results, you see the duality I was speaking about. Therefore, rather than asking what is the structure of $\mathcal{X}^-$ which, as seen above, is huge and morally free, one can wonder about the structure of $X^+$. There is an old-standing conjecture by Ralph Greenberg (who never, to my knowledge, baptized this question as "conjecture", but wrote the beautiful paper On the Iwasawa invariants of totally real number fields, American J. of Math, (1976), 263-284, where he studied it) predicting that $X^+\sim 0$, i. e. it is finite. Much work has been done on this (google-it to verify, being aware that Greenberg posed several conjectures in Iwasawa theory, so you should double-check the results) but it remains open and is probably one of the last big open question about the Iwasawa theory of totally real fields, together with Leopoldt.

Two last words: first of all, much of the above generalizes almost immediately to an abelian real base field $F$ insted of $\mathbb{Q}$ (just to be safe, suppose may be that $p\nmid [F:\mathbb{Q})$. Secondly, a nice table describing these dualities (and some more involving cyclotomic units) and their relations with different Main Conjectures is depicted on page 349 of Washington's book.

share|improve this answer

Thank you very much.

Kummer-Vandiver conjecture will imply $X^+=0$, not just $X^+\sim 0$. Recently, I found some connection between Kummer-Vandiver conjecture and the freeness of $\mathcal{X}^-$ as a $\mathbb{Z}p[[\Gamma]]$-module. That is why I asked about $\mathcal{X}^-$.

share|improve this answer
    
This seems to be more appropriate as a comment. –  S. Carnahan Apr 12 '13 at 3:20
    
I guess she/he has not enough reputation yet, one needs 50 points. –  Filippo Alberto Edoardo Apr 12 '13 at 5:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.