What's the class group of $\mathbb{Q}^{\mathrm{ab}}$?

Question

If $K/\mathbb{Q}$ is an infinite algebraic extension, define as usual the class group $Cl_K$ by the direct limit via the natural (conorm) map $Cl_K := \lim\limits_{\rightarrow} Cl_F$, where $F$ runs over all finite extensions of $\mathbb{Q}$. (I think this definition coincides with the Picard group of the ring of integers in $K$.)

For example, it is easy to see $Cl_{\overline{\mathbb{Q}}}=0$. From Iwasawa theory if $K$ is a $\mathbb{Z}_p$-extension of some finite extension of $\mathbb{Q}$, we know that the $p$-primary part of the class group (I will call it $p$-class group for short): $$Cl_K(p)=(\mathbb{Q}_p/\mathbb{Z}_p)^{\lambda} \times \text{ some p-group of bounded exponent}$$ If moreover $K$ is abelian over $\mathbb{Q}$, Ferrero-Washington proved that the $p$-group on the right hand side is $0$.

Now let $K=\mathbb{Q}(\mu_{\infty})$, the field joining all the roots of unity, which is the maximal abelian extension of $\mathbb{Q}$. What is $Cl_K$ or what is $Cl_K(p)$?

I think $Cl_K$ has subgroups of forms, for example $H=\mathbb{Q}_{37}/\mathbb{Z}_{37}$, because $H$ is the $37$-class group of $F:=\mathbb{Q}(\mu_{37^\infty})$ and one can show that the natural map from $Cl_{F}(p)$ to $Cl_K$ is injective, using the fact that $H$ is in the minus part (the complex conjugatation acts by $-1$) of $Cl_F$.

Answer 1

From Armand Brumer, The class group of all cyclotomic integers:

As an abelian group, $\mathrm{Pic}(O_\infty)$ is isomorphic to a countable direct sum of copies of $\mathbb Q/\mathbb Z$.

Here $O_\infty$ is the ring of cyclotomic integers, and $\mathrm{Pic}$ is the Picard group - for a number ring it coincides with the class group, while for a union of an increasing tower of such it coincides with the direct limit, as in your definition.