Let $k$ be a number field and let $k_\infty$ be the cyclotomic $\mathbf{Z}_p$-extension of $k$. Put $\Gamma=G(k_\infty/k)\cong \mathbf{Z}_p$, $\Lambda=\mathbf{Z}_p[[\Gamma]]$. Let $S$ be a finite set of primes of $k$ which contains primes above p and infinite primes. For simplicity, assume $p\neq 2$. Put

$L_\infty=$ the maximal unramified abelian pro-$p$-extension of $k_\infty$, $\quad$ $X=G(L_\infty/k_\infty)$;

$L_\infty^S=$ the maximal $S$-ramified abelian pro-$p$-extension of $k_\infty$, $\quad$ $X_S=G(L_\infty^S/k_\infty)$.

As usual, both $X$ and $X_S$ are finitely generated $\Lambda$-module ($X$ is even torsion). Thus one can talk about their $\mu$-invariants. My question is, is it true that $\mu(X)=\mu(X_S)$? (Or weaker: Is it true that $\mu(X)=0\Longleftrightarrow \mu(X_S)=0$?)

I know that when $k$ contains $\mu_p$ ($p$-th roots of unity), then $\mu(X)=\mu(X_S)$. (Ref. Cohomology of Number Fields, XI. $\S 3$). But what if $\mu_p\not\subseteq k$?

Another question. Suppose that $K/k$ be a finite extension, unramified outside $S$. Then it is easy to prove $\mu(X(k))\le \mu(X(K))$. (Ref. Washington's book (2nd Ed.), $\S 7.5$) Is it true that $\mu(X_S(k))\le \mu(X_S(K))$?


Yes, $\mu(X_S)=\mu(X)$, this was proved by Iwasawa in "On $\mathbb{Z}_l$-extensions of algebraic number fields". This is stated for instance as Theorem 2.5 in Sujatha's article "Elliptic Curves and Iwasawa's $\mu=0$ conjecture".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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