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Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_l]$ in $l$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_l]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_l) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_l) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $l$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_l)$ and $(x_1',\dotsc, x_l')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution (not sure) i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$, i.e, $\operatorname{Gal}(L'/L)$ acts transitively on $S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation (i.e., acts transitively) on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_l]$ in $l$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_l]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_l) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_l) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $l$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_l)$ and $(x_1',\dotsc, x_l')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_n]$ in $n$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_n]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_n) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_n) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $n$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_n)$ and $(x_1',\dotsc, x_n')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution (not sure) i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$, i.e, $\operatorname{Gal}(L'/L)$ acts transitively on $S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation (i.e., acts transitively) on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_l]$ in $l$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_l]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_l) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_l) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $l$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_l)$ and $(x_1',\dotsc, x_l')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution (not sure) i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$, i.e, $\operatorname{Gal}(L'/L)$ acts transitively on $S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation (i.e., acts transitively) on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_n]$ in $n$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_n]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_n) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_n) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $n$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a proper subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_n)$ and $(x_1',\dotsc, x_n')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution (not sure) i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$, i.e, $\operatorname{Gal}(L'/L)$ acts transitively on $S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation (i.e., acts transitively) on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_n]$ in $n$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_n]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_n) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_n) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $n$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

1. Is (or when is) the extension $L(S)/L$ Galois?

2. When is $L(S)/L$ totally ramified?

My Effort:

1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_n)$ and $(x_1',\dotsc, x_n')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution (not sure) i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$, i.e, $\operatorname{Gal}(L'/L)$ acts transitively on $S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation (i.e., acts transitively) on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.