List all intermediate fields which is a Galois extension of $\mathbb{Q}$ Let $K$ be the splitting field of $f(x)=x^3+2x-4$
List all fields $L$ with  $\mathbb{Q} \subset L \subset K$ for which $L$ is a Galois extension of $\mathbb{Q}$.
Things I know:
Since this is about intermediate fields, we can try to apply the Fundamental Theorem of Galois theory.
EDIT: Note that we can compute that the Galois group $Gal(K/ \mathbb{Q})$ is $S_3$
So the corresponding Galois groups for $L$ are the normal subgroups of $S_3$
From here I'm not sure how to continue to find $L$.
Any help will be appreciated!
 A: The polynomial is primitive( $\gcd$ of coefficients  is $1$ as a polynomial over $\Bbb{Z}[x]$) and has no integer roots (it is easy to prove by elementary calculus or by drawing a graph) . So it is irreducible in $\Bbb{Z}[x]$ and hence also in $\Bbb{Q}[x]$ due to Gauss Lemma.
Also from elementary calculus, it is obvious that it has a unique real root (function is monotonically increasing) . Thus it has a unique real root and two complex roots which occur as conjugates.
Let $\alpha$ be it's unique real root.
So we have $[\Bbb{Q}(\alpha):\Bbb{Q}]=3$ and over $\Bbb{Q}(\alpha)$ , the polynomial factors into $(x-\alpha)(x-z)(x-\bar{z})$ . Where $z$ and $\bar{z}$ are the complex roots.
It is obvious that over $\Bbb{Q}(\alpha)$ the polynomial $(x-z)(x-\bar{z})$ is irreducible and hence $[\Bbb{Q}(\alpha,z):\Bbb{Q}(\alpha)]=2$ . And since quadratic extensions of $\Bbb{Q}$ are always Galois(alternatively you can prove that any quadratic extension of $\Bbb{Q}$ is a splitting field of a separable polynomial), you have that over $\Bbb{Q}(\alpha,z)$ our polynomial splits and it is the Splitting field as it is the smallest field containing $\alpha,z$.
Thus we have that $|\text{Gal}( \Bbb{Q}(\alpha,z)/\Bbb{Q})|=[\Bbb{Q}(\alpha,z):\Bbb{Q}]=3\cdot 2 = 6$ due to the fact that the compositum of two extensions of degrees $m,n$ such that $\gcd(m,n)=1$ has degree $m\cdot n$ .
Now it remains to prove that this group is indeed non-abelian . Take the elements $\sigma,\tau\in\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})$ such that $\sigma=\begin{cases}\alpha\mapsto z\\ z\mapsto \bar{z}\\\bar{z}\mapsto \alpha\end{cases}$ and $\tau=\begin{cases} z\mapsto \bar{z}\\ \bar{z}\mapsto z\\ \alpha\mapsto \alpha\end{cases}$ . Then we can say that $\sigma\cdot\tau\neq \tau\cdot\sigma $ which proves that the Galois group is non-abelian and as it is of order $6$ it must be $S_{3}$ .
Now the only proper normal subgroup of $S_{3}$ is the unique subgroup of order $3$. Here it is seen that $\sigma\in\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})$ does have order $3$. and hence the only such Galois extension $L$ of $\Bbb{Q}$ is the fixed field of $\langle \sigma\rangle $ and it is such that $[L:\Bbb{Q}]=\bigg|\frac{\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})}{\langle \sigma\rangle}\bigg|=\frac{6}{3}=2$ .
So $L=K^{\langle \sigma\rangle}$. That is the fixed field of $\sigma$.
More explicitly $K=\Bbb{Q}(\alpha,\sqrt{D})$ where $D$ is the discriminant of the polynomial. Look at Dummit and Foote page $613$ for reference. And hence $L=\Bbb{Q}(\sqrt{D})$ which is the unique degree $2$ extension of $\Bbb{Q}$ contained in $K$.
A short explanation regarding the above. We have $\displaystyle \prod_{i<j} (\alpha_{i}-\alpha_{j})^{2}=D $ and define $\sqrt{D}= \prod_{i<j} (\alpha_{i}-\alpha_{j})$ where $\alpha_{i}$'s are roots of the polynomial . Note that $\sqrt{D}$ is just a symbol for the expression. For $\sigma\in S_{n}$ you have $\sigma(\sqrt{D})=\text{sgn}(\sigma)\sqrt{D}$ where $\text{sgn}$ is the signum of the permutation.
Hence it is clear from above that $\sigma\in A_{n}\subset S_{n}$ if and only if $\sigma(\sqrt{D})=\sqrt{D}$ . Thus in our above case $A_{3}$ will fix $\Bbb{Q}(\sqrt{D})$ and it will be the unique quadratic extension of $\Bbb{Q}$ properly contained in $K$ .
Now as to how to compute the discriminant , you should take a look at page $613$ of Dummit and Foote.
A: The Galois group of your polynomial is not $\mathbb{Z}_{2}$, it's the symmetric group of order 6, $\mathbb{S}_3$. This is because the polynomial is irreducible in $\mathbb{Q}$ since it has no roots in the field and has degree $3$. Also the discriminant is $-464$, whose square root is not an element of the field.
Now if you want to know the list of subfields $L$ such $L|\mathbb{Q}$ is Galois, under the fundamental Theorem of Galois theory you just need to find the normal subgroups of the symmetric group of order 6
