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If K is a finite extension of $\mathbb Q_p$ for some prime number $p$, (possibly need $p \neq 2$), $L_1$ and $L_2$ are totally ramified abelian extension of $K$, $\pi_1, \pi_2$ are respectively the uniformizer that generates each field. Is it true that $L_1 L_2$ is totally ramified iff $Nm_{L_1 / K}(\pi_1)$, $Nm_{L_2 / K}(\pi_2)$ differs by an element in the intersection of the two norm groups.

Is there a proof of this result (or the correct version of the result) without employing big tools?

Add at 6:49 pm 18th Feb: From Class Field Theory we know that there are one maximal totally ramified abelian extension of a local number field $K$ corresponding to each uniformizer, so I would expect that some version of the above statement is true. At least when $Nm_{L_1 / K}(\pi_1)=Nm_{L_2 / K}(\pi_2)=x$, $L_1 L_2$ is totally ramified, because both are contained in $K^{ram}_x$.

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When does the composition of two totally ramified extension totally ramified?

If $L_1$ and $L_2$ are totally ramified abelian extension of $K$, $\pi_1, \pi_2$ are respectively the uniformizer that generates each field. Is it true that $L_1 L_2$ is totally ramified iff $Nm_{L_1 / K}(\pi_1)$, $Nm_{L_2 / K}(\pi_2)$ differs by an element in the intersection of the two norm groups.

Is there a proof of this result (or the correct version of the result) without employing big tools?
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