Let $M = \text{Aut}(\mathbb{Q}(i, \sqrt{2})) \cong V_{4}$ and show that the $\prod_{\sigma \in M}(X - \sigma(\alpha))$ is equal to the minimal polynomial of $\alpha = 1 + i + \sqrt{2}$ over $\mathbb{Q}$.
My attempt:
In order to show that $M \cong V_{4}$ it suffices to show that $M$ is of order $4$ and not cyclic. Note that under automorphism, roots of a polynomial are mapped to other roots of said polynomial. Note that the automorphisms in $M$ fix the prime field $\mathbb{Q}$. Then $M$ consists of
- The identity,
- The mapping that fixes $\sqrt{2}$ and maps $i$ to $-i$,
- The mapping that fixes $i$ and maps $\sqrt{2}$ to $-\sqrt{2}$,
- The mapping that maps $\sqrt{2}$ to $-\sqrt{2}$ and $i$ to $-i$.
This clearly resembles $V_{4}$, but another argument can be made. Note that $\mathbb{Z}/4\mathbb{Z}$ has an element of order $4$, while the automorphisms in $M$ (other than the identity of course) have order $2$. Then $M$ cannot be cyclic, hence $M \cong V_{4}$.
Now let $\alpha = 1 + i + \sqrt{2}$, we want to show that $f = \prod_{\sigma \in M}(X - \sigma(\alpha)) \in \mathbb{Q}[X]$ is equal to the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Clearly, $\alpha$ is a root of $f$. Then $f^{\alpha}_{\mathbb{Q}}$ divides $f$. Because of the mapping of roots to other roots under $\sigma \in M$, every root $\sigma(\alpha)$ of $f$ is also a root of the minimal polynomial.
I am not sure whether this is the right approach, and whether this is enough to conclude that $f$ and the minimal polynomial are equal in $\mathbb{Q}$. I am unsure how to account for multiplicity as well.
I am quite sure about the first part of the proof. Would someone be able to help me with the second part? Thanks in advance. I am not familiar with Galois groups and extensions yet.
Your approach is basically correct. To "complete" the second part, you could note that the $\sigma(\alpha)$ are all distinct, so the fact that $M$ acts transitively on the roots of $f$ implies that it is irreducible, hence the minimal polynomial of $\alpha$.
Your reasoning in the first part is correct but somewhat informal (with regards to how you determine the members of $M$), so let me give a quick, more rigorous argument for that part as well: Clearly $\newcommand{\Q}{\mathbb{Q}}\Q(\sqrt{2}) / \Q$ is a quadratic extension, so its Galois group is isomorphic to $\mathbb{Z} / 2$ with the nontrivial element given by $\sqrt{2} \mapsto -\sqrt{2}$ by the extension theorem. Now $\Q(\sqrt{2}, i) / \Q(\sqrt{2})$ is again quadratic with unique nontrivial $\Q(\sqrt{2})$-automorphism given by $i \mapsto -i$, so putting these together (again using the extensions theorem) you obtain the description of $M$ you gave.