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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 $\bar K^\times \overline{K^\times} = K^\times/K^{\times p}$ (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 $L=K(\root p\of D)$) 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 $\bar K=K/\wp(K)$ \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$-line$D$\bar{\mathfrak o}$ (say).

Thus, the compositum of two totally ramified abelian extensions $L_i|K$ of exponent $p$ is totally ramified precisely when the subspace $D_1D_2$ D_1+D_2$does not contain the line$D$, \bar{\mathfrak o}$, where $D_i$ is the subspace giving rise to $L_i$. L_i$in the sense that$L_i=K(\wp^{-1}(D_i))$. See Parts 5 and 6 of arXiv:0909.2541. 4 Addendum. Let me give an elementary answer in the case of abelian exponent-$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$\bar K^\times = K^\times/K^{\times p}$(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$L=K(\root p\of D)$) 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$\bar K=K/\wp(K)$(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$D$(say). Thus, the compositum of two totally ramified abelian extensions$L_i|K$of exponent$p$is totally ramified precisely when the subspace$D_1D_2$does not contain the line$D$, where$D_i$is the subspace giving rise to$L_i$. See Parts 5 and 6 of arXiv:0909.2541. 3 added 153 characters in body 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$\bar K^\times = K^\times/K^{\times p}$(thought of a vector space over$\mathbb{F}_p$). \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 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 $p$ 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 $\bar U_{pe_1}$.

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