If K is a finite extension of $\mathbb Q\_p$ for some prime number $p$, (possibly need $p \neq 2$), $L\_1$ and $L\_2$ 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 $K^{ram}\_x$.



When is the composition of two totally ramified extension totally ramified?
Let me give an elementary answer in the case of abelian exponent-$p$ extensions of $K$, where $K$ is a finite extension of $\mathbb{Q}\_p$ containing a primitive $p$-th root $\zeta$ of $1$. This is the basic case, and Kummer theory suffices.


Such extensions correspond to sub-$\mathbb{F}\_p$-spaces in $\overline{K^\times} = K^\times/K^{\times p}$ (thought of a vector space over $\mathbb{F}\_p$; not to be confused with the multiplicative group of an algebraic closure of $K$). 


It can be shown fairly easily that the unramified degree-$p$ extension of $K$ corresponds to the $\mathbb{F}\_p$-line $\bar U\_{pe\_1}$, where $e\_1$ is the ramification index of $K|\mathbb{Q}\_p(\zeta)$ and $\bar U\_{pe\_1}$ is the image in $\bar K^\times$ of the group of units congruent to $1$ modulo the maximal ideal to the exponent $pe\_1$. This is the "deepest line" in the filtration on $\bar K^\times$. See for example prop. 16 of [arXiv:0711.3878](http://arxiv.org/abs/0711.3878v2).


An abelian extension $L|K$ of exponent $p$ is totally ramified if and only if the subspace $D$ which gives rise to $L$ (in the sense that $L=K(\root p\of D)$) does not contain the line $\bar U\_{pe\_1}$.


Now, if $L\_1$ and $L\_2$ are given by the sub-$\mathbb{F}\_p$-spaces $D\_1$ and $D\_2$, then the compositum $L\_1L\_2$ is given by the subspace $D\_1D\_2$ (the subspace generated by the union of $D\_1$ and $D\_2$). Thus the compositum $L\_1L\_2$ is totally ramified if and only if $D\_1D\_2$ does not contain the deepest line $\bar U\_{pe\_1}$.


**Addendum.** A similar remark can be made when the base field $K$ is a finite extension of $\mathbb{F}\_p((\pi))$. Abelian extensions $L|K$ of exponent $p$ correspond to sub-$\mathbb{F}\_p$-spaces of $\overline{K^+}=K/\wp(K^+)$ (not to be confused with an algebraic closure of $K$), by Artin-Schreier theory. The unramified degree-$p$ extension corresponds to the image of $\mathfrak{o}$ in $\bar K$, which is an $\mathbb{F}\_p$-line $\bar{\mathfrak o}$ (say).


Thus, the compositum of two totally ramified abelian extensions $L\_i|K$ of exponent $p$ is totally ramified precisely when the subspace $D\_1+D\_2$ does not contain the line $\bar{\mathfrak o}$, where $D\_i$ is the subspace giving rise to $L\_i$ in the sense that $L\_i=K(\wp^{-1}(D\_i))$. See Parts 5 and 6 of [arXiv:0909.2541](http://arxiv.org/abs/0909.2541v2).


