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

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

1 Answer 1

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

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    $\begingroup$ 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? $\endgroup$
    – user755024
    Commented Oct 3, 2020 at 4:25
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    $\begingroup$ @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 $\endgroup$
    – N. S.
    Commented Oct 3, 2020 at 20:18
  • $\begingroup$ $$a^4-2a^2+2=0$$ for some $a \in \mathbb Q$. But this is irreducible over $\mathbb Q$ by Eisenstein. $\endgroup$
    – N. S.
    Commented Oct 3, 2020 at 20:19
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    $\begingroup$ Note that when you get $\beta^2=2+2\sqrt{2}$ you can probably use norms to finish. $\endgroup$
    – N. S.
    Commented Oct 3, 2020 at 20:19

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