Cubic extension and Galois theory [closed]

Q1: If $K \cong \sigma(K)$, this isomorphism must give you a map $K \to \sigma(K) \subset \overline{\mathbb{Q}}$ automatically

Q2: That is not true. You can see a degree $3$ Galois extension here

If what you mean is that if the galois closure has degree $6$, then the Galois group is $S_3$, then that is true. This is because if $K = \mathbb{Q}(\theta)$ (which is possible by the theorem of the primitive element), then the Galois closure is the splitting extension of the minimal polynomial of $\theta$, which must be of degree $3$ because it is a cubic extension. But then the Galois group acts on the roots by permuting them (and any automorphism of the field is determined by the action on these roots so they generate the field) so it must be a subgroup of $S_3$.

Finally, $\text{Gal}(K^\text{Gal}| \mathbb{Q}) = [K^\text{Gal}: \mathbb{Q}]$ and $S_3$ is its only subgroup with $6$ elements.

Q3: Yes, this is true. Normality gives you that $\sigma (\theta) \in K$, since $\theta$ must be a root of the minimal polynomial of $\theta$ (roots are preserved by $\mathbb{Q}$-automorphisms). Then, $\deg m_\theta (x) = [K : \mathbb{Q}] = [\mathbb{Q}(\sigma(\theta)): \mathbb{Q}]$. Since it is a subextension and it has the same degree it must be equal.

Q4: Yes, they are both isomorphism to $\mathbb{Q}[x]/m_\theta(x)$.

Q5: Yes, this is an easy consequence of the fundamental theorem of Galois theory.

Q6: Yes