Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$. Let us consider the polynomial ring $R=K[x_1,x_2,...,x_n]$ 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:
$$S=\bigcup_{1 \leq i \leq m}\{(x_1, \cdots, x_n)~|~f_i(x_1, \cdots, x_n)=0 \},$$
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 $L$. 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(a_{i1}, \cdots,a_{in})=0}L(a_{i1},\cdots, a_{in}), \ i=1,\cdots, m; \\
&=L({\color{red}{a_{11},\cdots, a_{1n}}}, {\color{blue}{a_{21},\cdots, a_{2n}}}, \cdots,{\color{green}{ a_{m1},\cdots,a_{mn}}}).
\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 ?

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**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 an 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. In our case $L(S)$ is generated by the $mn$ elements $a_{ij}, \ 1 \leq i \leq m, \ 1 \leq j \leq n$. So $L(S)$ will be totally ramified over $L$ if $L(S)=L(a_{ij})$, where $a_{ij}$ is a root of an Eisenstein polynomial over $L$. These are all my intuition. Is there other argument ?


Any discussion please