The ideal class group $\mathrm{Cl}({\cal O}_K)$ and Mazur's deformation ring $R(\overline{\rho})$ for a number field $K$ are said to be similar to each other.

Let ${\Bbb Q}_{\infty}$ be the unique cyclotomic ${\Bbb Z}_p$ extension of $\Bbb Q$ and ${\Bbb Q}_n$ be its $n$-th layer, i.e. ${\mathrm{Gal}}({\Bbb Q}_n/{\Bbb Q}) \cong {\Bbb Z}/p^n{\Bbb Z}$. Beginning from the fixed absolutely irreducible Galois representation $\overline{\rho}_0 \colon G_{\Bbb Q} \to {\mathrm{GL}}_2({\Bbb F}_p)$, we consider its lift to each ${\Bbb Q}_n$ and get $\overline{\rho}_n \colon G_{{\Bbb Q}_n} \to {\mathrm{GL}}_2({\Bbb F}_p)$.

On the other hand, we can associate ideal class group $\mathrm{Cl}({\cal O}_{{\Bbb Q}_n})$ and Mazur's deformation ring $R(\overline{\rho}_n)$ for each ${\Bbb Q}_n$. We take the inverse limits of these as follows:

$\mathrm{Cl}^{\infty} \colon\!= \varprojlim_{n}\mathrm{Cl}({\cal O}_{{\Bbb Q}_n})\{p\} \cong \underset{i=1 \ldots r}{\bigoplus} \Lambda/p^{\mu_i} \oplus \underset{i=1 \ldots s}{\bigoplus} \Lambda/f_j(T)^{m_j}$, where $\Lambda \colon= {\Bbb Z}_p[[T]]$ is the Iwasawa algebra

$R^{\infty} \colon= \varprojlim_{n}R(\overline{\rho}_n) \cong {\Bbb Z}_p[[X_1,...,X_d]]/J$.

Question: Why does the condition $d < \infty$ for $R^{\infty}$ correspond to $\mu = \sum_{i=1 \ldots r} \mu_i =0$ for $\mathrm{Cl}^{\infty}$?

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    $\begingroup$ This question is quite confused (among many other things, $\bar{\rho}_{n}$ might not remain absolutely irreducible). Also, asking why an object corresponds to another is not very precise (I'm guessing that you have read this somewhere and you are trying to understand, but then you should tell us where, so that we can judge the context). That said, your $d$ is the dimension of a Galois cohomology group attached to the adjoint representation of $\bar{\rho}$, so its finiteness is related to finiteness of Selmer groups, chief among them finiteness for the class groups. $\endgroup$ – Olivier Dec 16 '13 at 10:53
  • $\begingroup$ Olivier, thanks. The book is Hilbert Modular Forms and Iwasawa Theory (Oxford Mathematical Monographs). $\endgroup$ – Pierre Dec 16 '13 at 11:04
  • $\begingroup$ Olivier, you said that chief among them finiteness for the class groups. So then is it obvious that finiteness for the class groups <=> mu = 0? $\endgroup$ – Pierre Dec 16 '13 at 11:13
  • $\begingroup$ No it is not obvious. $\endgroup$ – Olivier Dec 16 '13 at 13:22

First of all, let us assume $p\neq 2$, otherwise I think $\bar{\rho}_n$ might be (absolutely) reducible even though $\bar{\rho}$ is not. Also, one has to make clear that the deformations parametrized by $R(\bar{\rho}_n)$ are unramified outside a finite sets of finite places of $\mathbb Q_n$ above a fixed finite sets of places $S$ of $\mathbb Q$ (or in other words that the representations considered are continuous representations of $G_{\mathbb Q,S}$.

Under these (perhaps implicit) assumptions, $R^\infty$ is well-defined, and the usual computation (originally due to B.Mazur) identifies its tangent space with $H^1(G_{\mathbb Q,S},\operatorname{ad}_{\bar{\rho}})$. The ring $R^{\infty}$ is noetherian (if and) only if its (co)tangent space is finite dimensional as a vector space over $\mathbb F_{p}$, so the fact that $R^\infty$ is noetherian or not amounts to whether the cohomology group $H^1(G_{\mathbb Q,S},\operatorname{ad}_{\bar{\rho}})$ is finite-dimensional over $\mathbb F_p$ or not. Now finally, a torsion-module over an Iwasawa algebra is of finite-dimensionality over $\mathbb F_p$ if and only if the $\mu$-invariant of its characteristic series is 0. In that sense, the notions you introduce correspond.

The analogy can be made closer by noticing that the class group of the ring of integers of an abelian extension $F$ of $\mathbb Q$ can indeed be expressed as a Galois cohomology group with coefficients in an adjoint representation (which in that case is simply the induction of the trivial representation of $G_F$ to $G_\mathbb Q$).

  • $\begingroup$ @Pierre Well, if you are happy with the answer, you can accept it, so that the software registers it correctly. $\endgroup$ – Olivier Dec 17 '13 at 9:38

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