Is $\Bbb{Z}_p$ or $\Bbb{Z}_p^2$ extension unramified

algebraic-number-theoryclass-field-theorygalois-theoryp-adic-number-theory

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).

Best Answer

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.

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