# Prove that $X^4 - 2X^2+ 2$ is the minimial polynomial of $\sqrt{1 + i}$ over $\mathbb{Q}(\sqrt{2})$.

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 $$\mathbb{Q}(\sqrt{2})$$?

• Could you correct the typo in the title? Commented Oct 3, 2020 at 3:48
• @Lubin Done, thank you for pointing it out.
– user755024
Commented Oct 3, 2020 at 3:51
• I probably have the ability to do that for others, but I think it’s less rude to make the request. Thanks. Commented Oct 3, 2020 at 3:53

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 $$\sqrt{2} \in \mathbb Q(\alpha)$$.

• Great answer, thank you for that. Let $N$ be the field norm of $\mathbb{Q}(\alpha)/\mathbb{Q}$. If $\sqrt{2} \in \mathbb{Q}(\alpha)$, then $N(2) =\text{det}\left(2 \ \text{id}_{\mathbb{Q}(\alpha)}\right) = 2$ and $N(2) = N(\sqrt{2})^2$, since the field norm is multiplicative. Moreover since $N$ takes values in $\mathbb{Q}$, this would imply that $2$ is the square of an element of $\mathbb{Q}$, which is absurd. Is there a simpler way?
– user755024
Commented Oct 3, 2020 at 4:25
• @EhWha Unless I am doing a stupid mistake, the following works. Assume by contradiction that $\sqrt{2} \in \mathbb Q( \alpha)$. Then $$\mathbb Q(i, \sqrt{2}) \subseteq \mathbb Q(\alpha)$$ and hence $$\mathbb Q(i, \sqrt{2}) = \mathbb Q(\alpha)$$ Next, $\alpha \bar{\alpha} =\sqrt{2}$ and hence $$\bar{\alpha} \in \mathbb Q(i, \sqrt{2})=\mathbb Q(\alpha)$$ Then $$\beta:=\alpha +\bar{\alpha} \in Q(i, \sqrt{2}) \cap \mathbb R =\mathbb Q(\sqrt{2})$$ and $$\beta^2= \alpha^2+\bar{\alpha}^2+2\alpha\bar{\alpha}=(1+i)+(1-i)+2 \sqrt{2}= 2+2\sqrt{2}$$ Now, writing $\beta=a+b\sqrt{2}$ and expanding you get Commented Oct 3, 2020 at 20:18
• $$a^4-2a^2+2=0$$ for some $a \in \mathbb Q$. But this is irreducible over $\mathbb Q$ by Eisenstein. Commented Oct 3, 2020 at 20:19
• Note that when you get $\beta^2=2+2\sqrt{2}$ you can probably use norms to finish. Commented Oct 3, 2020 at 20:19