Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$. Let us consider the polynomial ring $K[x_1,x_2,...,x_n]$ in $n$-variables and $f_1, f_2, \cdots, f_m \in K[x_1, \cdots, x_n]$.

Consider a finite extension $L$ of $K$ and consider the following zero set: $$S=\bigcup_{1 \leq i \leq m}\{(x_1, \cdots, x_n) \in \bar L^n~|~f_i(x_1, \cdots, x_n)=0 \},$$ where each $f_i \in L[x_1,x_2, \cdots , x_n]$. Then, clearly every element in $S$ is a point in an affine $n$-space over $\bar L$. So the coordinates of each points in $S$ generate a field extension. i.e., conisder the field extension $L(S)$ obtained adjoining the coordinates of each solutions in $S$. That is, \begin{align} L(S)&=\bigcup_{f_i(x_1, \cdots,x_n)=0}L(x_1,\cdots, x_n), \ i=1,\cdots, m; \end{align}

Questions:

$(1)$ Is (or When) the extension $L(S)/K$ Galois ?

$(2)$ Is (or When) the extension $L(S)/L$ Galois ?

$(3)$ When are the above two extensions totally ramified ?

$$-----------$$ My effort:

$(1)$ If we assume $L$ is an unramified extension of $K$, then $L/K$ is Galois extension. Now $L(S)$ is the algebraic extension of $L$ because its elements are algebraic over $L$. Thus $L(S)/L$ is also Galois extension. Hence $L(S)/K$ is Galois extension. In this multivariable case, we don't need separability of the roots because we are taking the coordinates only. Also two solutions $(x_1, \cdots,x_n)$ and $(x_1',\cdots, x_n')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both gives the same extension, so we will take just one of the coordinates. Am I correct ?

$(2)$ Again if we assume $L$ is unramified extension of $K$, then by the same argument $L(S)/L$ is Galois. Is there argument if we don't assume that '$L$ is unramified' ?

$(3)$ It is not clear to me. But in single variable case, we say that a finite Galois extension $L$ of $K$ is totally ramified if $L$ is the simple extension by a root $a$, say, of an Eisenstein polynomial. i.e., $L=K(a)$, where $a$ is a root of an Eisenstein polynomial. So $L(S)$ will be totally ramified over $L$ if $L(S)=L(x_i)$, where $x_i$ is a root of an Eisenstein polynomial over $L$.

There are other equivalent two arguments as well. $(i)$ If $Gal(L(S)/L)$ coincides with its inertia subgroup, then it will be totally ramified. $(ii)$ If the norm $N(L(S)/L)$ contains an uniformizer of $L$, then it will be totally ramified. These are all I can think about but couldn't answer the questions.

Any discussion please