For fixed prime $l$, let $K$ be a finite extension of $ \Bbb{Q}_l$, and $L$ be algebraic extension of $K$. For $p≠l$,if $Gal(L/K)$ is isomorphic to $\Bbb{Z}_p$ or $\Bbb{Z}_p^2$, my pdf reads from class field theory, such extension $L/K$ is unramified.
But what is exactly the proposition we used from class field theory in this situation? Thank you for your help.
cf. My pdf is 'Elliptic curves with complex multiplication and the conjecture of birch Swinnerton dyer' by Rubin. Theorem 5.7,(ⅱ).
On
I'm answering to the orignal question where $p=l$, the case $p\ne l$ is very different and is answered by Lukas below.
This is not true as $\Bbb{Q}_3(\zeta_{3^\infty})/\Bbb{Q}_3(\zeta_3)$ is totally ramified with Galois group $$Gal(\Bbb{Q}_3(\zeta_{3^\infty})/\Bbb{Q}_3(\zeta_3)) \cong 1+3\Bbb{Z}_3\cong \Bbb{Z}_3$$ An unramified $\Bbb{Z}_p$-extension always exists and is unique (can you construct it?) and an unramified $\Bbb{Z}_p^2$-extension never exists.
reuns answered the original question. I‘ll answer the revised question.
The local reciprocity map furnishes an isomorphism $\widehat{K^\times}\xrightarrow{\cong}G^{ab}$, where the hat denotes profinite completion and $G$ is the absolute Galois group of $K$.
Now since $K^\times\cong\pi^\Bbb Z \times \mathscr O_K^\times$, we have a canonical (after choosing a uniformiser $\pi$) isomorphism $$\widehat{K^\times}\cong \widehat{\Bbb Z}\times\mathscr O_K^\times$$ By local class field theory, under this isomorphism the first factor corresponds to the Galois group of the maximal unramified extension. Thus, by infinite Galois theory, it suffices to show that every continuous homomorphism $\mathscr O_K^\times \to \Bbb Z_p$ is trivial. This is not too difficult: by the structure theorem for $\mathscr O_K^\times$, we know that it is a product of a finite group and a pro-$\ell$ group. But for $\ell \neq p$, the only continuous homomorphism from a pro-$\ell$ group to a pro-$p$ group is trivial. And a homomorphism from a finite group to a torsion-free group is also trivial.