Example of non Galois extension of a local field and its Galois closure I am trying to think of an example of non Galois extension of a local field and its Galois closure.
I started by looking at examples of extensions of $\mathbb Q_p $.
For example

*

*I think $x^3-3$ is irreducible over $\mathbb Q _3 $ but I do not know how to prove that. I think I should prove that there is no root of the polynomial in $\mathbb Q_3$.
(I tried to used Sagemath with commands
R. = Qp(3,print_mode='digits')[]
f = x^3 - 3
f.is_irreducible()
And got output True.)  How do I find the Galois closure ?


*$x^3+2x^2+2x+2$ is irreducible over $\mathbb Q_3 $ as it does not have any root in $\mathbb Z /3 \mathbb Z $. But even finding its Galois group seems difficult.


Motivation:
I want to see Norm limitation theorem and its proof in action. The norm limitation theorem says that for any extension $L$ over a local field $K$, $Nm(L)$ =$Nm(E)$ where $E$ is maximal abelian extension in $L$ over $K$. In proof of the theorem, the key step is to consider a galois extension containing $L$ and consider a commutative diagram.
So I was looking above for example for this.
 A: The Galois group of an irreducible cubic $f(x)$ over a field $k$ of characteristic $0$ is $S_3$ if the discriminant $\Delta$ isn't a square and $A_3$ if it is (see, for example, Keith Conrad's Galois groups of cubics and quartics, Theorem 2.1). In the former case $L = k[x]/f(x)$ will be a non-Galois extension (because it has degree $3$ but the Galois group has degree $6$) and its Galois closure will be a quadratic extension of it.
$x^3 - 3$ is irreducible over $\mathbb{Q}_3$ because any root would have to have $3$-adic valuation $\frac{1}{3}$. Its discriminant is $-3^5$ which is not a square because it has odd $3$-adic valuation, so the Galois group is $S_3$ and $\mathbb{Q}_3[x]/(x^3 - 3)$ is a non-Galois extension as desired.
$x^3 + 2x^2 + 2x + 2$ has discriminant $-44 \equiv 1 \bmod 3$ which is a square in $\mathbb{Q}_3$ by Hensel's lemma so its Galois group is $A_3$.
In general for $p$ an odd prime a nonzero element of $\mathbb{Q}_p$ is a square iff it has the form $p^{2k} u$ where $k \in \mathbb{Z}$, $u \in \mathbb{Z}_p^{\times}$ is a unit, and $u$ is a square $\bmod p$ (by Hensel's lemma again).
