Does $\mu=0$ for an imaginary quadratic field $K$ imply $\mu=0$ for $\mathbf{Q}$? Suppose that $E/\mathbf{Q}$ is an elliptic curve and $K$ is an imaginary quadratic field. Let $\mathbf{Q}_{\infty}$ denote the cyclotomic $\mathbf{Z}_p$ extension of $\mathbf{Q}$, and let $K_{\infty}$ denote the cyclotomic $\mathbf{Z}_p$ extension of $K$,
If we know that the $\mu$-invariant for the Selmer over $K_{\infty}$ vanishes, does it follow that the $\mu$-invariant for the Selmer over  $\mathbf{Q}_{\infty}$ vanishes? That is, if we know $\mu=0$ for the bigger field $K_{\infty}$, can we "propogate it down" to show that $\mu=0$ for the field $\mathbf{Q}_{\infty}$?
 A: Under your setting
\begin{equation}
\operatorname{Sel}(E/K_{\infty})\simeq\operatorname{Sel}(E/\mathbb  Q_{\infty})\oplus\operatorname{Sel}(E\otimes\chi/\mathbb Q_{\infty})
\end{equation}
where $\chi$ is the quadratic character attached to $K/\mathbb Q$ and all these $\mathbb Z_{p}[[X]]$-modules are torsion. Hence
\begin{equation}
\mu(\operatorname{Sel}(E/K_{\infty}))=\mu(\operatorname{Sel}(E/\mathbb  Q_{\infty}))+\mu(\operatorname{Sel}(E\otimes\chi/\mathbb Q_{\infty})).
\end{equation}
Indeed, if $\mu(\operatorname{Sel}(E/K_{\infty}))=0$ then $\mu(\operatorname{Sel}(E/\mathbb  Q_{\infty}))=0$.
However, the very same argument suggests that it is probably very hard to show that $\mu(\operatorname{Sel}(E/K_{\infty}))=0$ without showing that $\mu(\operatorname{Sel}(E/\mathbb  Q_{\infty}))=0$.
What has been known since the works Gillard, Schneps, Hida, Hsieh... is that it is sometimes possible to prove that $\mu$-invariant vanishes over the $\mathbb Z_p^2$-extension $L_\infty/K$ (the composite of the cyclotomic and anticyclotomic extension of $K$). Unfortunately, in that case, there is no obvious way (and indeed, at present no known way) to deduce from those results the vanishing of the $\mu$-invariant over $\mathbb Q_\infty$, as the specialization of a power-series in $\mathbb Z_p[[X,Y]]$ with vanishing $\mu$-invariant at $Y=0$ can very well have non-vanishing $\mu$-invariant.
