algebraic number theory - 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.

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 $L=K(root pof D)$) 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.

• Next nt.number theory - Composite pairs of the form n!-1 and n!+1
• Previous ag.algebraic geometry - Hilbert scheme of points on a complex surface