Context
Let $\alpha$ be a square root of $1 + i$, and define a polynomial $p := X^4 – 2X^2 + 2 \in \mathbb{Q}(\sqrt{2})[X]$. As the title states, I am wondering how to show that $p$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$. This question has come about while I have been solving a problem which requires me to find the degree of the extension $\mathbb{Q}(\sqrt{2}, \alpha)/\mathbb{Q}$. My approach is to use the tower law: $[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]$. I know that $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}] = 2$, so the problem boils down to finding $[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\sqrt{2})]$, which is just the degree of the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$.
Partial solution
I know that $p$ is the minimial polynomial of $\alpha$ over $\mathbb{Q}$ since $X^4 – 2X^2 + 2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion, and $\alpha$ is a root of $p$.
I suspect that $p$ is also the minimal polynomial for $\alpha$ over $\mathbb{Q}(\sqrt{2})$, and this is my (not completely rigorous) reasoning.
If we start with $\alpha = \sqrt{1 + i}$, then by repeatedly squaring until we end up with an element of $\mathbb{Q}(\sqrt{2})[\alpha]$, we get $(\alpha^2 -1)^2 = -1$, or $\alpha^4 – 2\alpha^2 + 2 = 0$. In similar problems, I usually conclude that we have found the required minimal polynomial, in this case the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$, being $X^4 – 2X^2 + 2$. The reason being that we found that $\alpha$ is a root of $X^4 – 2 X^2 + 2 \in \mathbb{Q}[X]$ by squaring $\alpha = \sqrt{1 + i}$ just enough times, though this is not satisfactory.
My question is this:
How could one formally argue that $X^4 – 2 X^2 + 2$ is indeed the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$?
Best Answer
You know that
$$[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}] =2 [\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\sqrt{2})] \\ [\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\sqrt{2})] \leq 4$$
Now, look at $$[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha) : \mathbb{Q}]= 4 [\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\alpha)]$$
From here you can decide that $[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\alpha)] \in \{ 1,2 \}$.
Therefore, either $\mathbb{Q}(\sqrt{2}, \alpha)= \mathbb{Q}(\alpha)$ or $[\mathbb{Q}(\sqrt{2}, \alpha): \mathbb{Q}(\sqrt{2})] =4$.
Now try to see if you can prove/disprove $\sqrt{2} \in \mathbb Q(\alpha)$.