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We have, $|\mathbb{Q}(i) : \mathbb{Q}(\sqrt{2})| = 2$ because a basis for this is $1$ and $i$ and $|\mathbb{Q}(\sqrt{2}) : \mathbb{Q}|$ = 2 since $1, \sqrt{2}$ is a basis. So shouldn't we get that $|\mathbb{Q}(i, \sqrt{2}): \mathbb{Q}| = 4$?

Context: I am working on a problem that asks to show $\mathbb{Q}(i, \sqrt{2}) = \mathbb{Q}(\alpha)$ for $\alpha$ an eighth root of unity. So the minimal polynomial for $\alpha$ would be $x^8-1$ over $\mathbb{Q},$ making the extension of $\alpha$ over $\mathbb{Q}$ degree $8.$ To show that the two extensions fields are equal I need only show that they have the same degree, but I am running in to some trouble here. Is this not the correct minimal polynomial? I know it reduces, so is it possible that $\alpha$ satisfies $x^4+1$ or $x^2+1$?

Mike
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    Recall: Minimal polynomials are irreducible. – Qi Zhu Jan 17 '22 at 16:45
  • In general, if $\alpha$ is a primitive $n$th root of unity, then $[\mathbb{Q}(\alpha) : \mathbb{Q}] = \varphi(n)$ where $\varphi$ is the Euler phi function. And the irreducible polynomial of $\alpha$ over $\mathbb{Q}$ is a cyclotomic polynomial. – Daniel Schepler Jan 17 '22 at 17:04
  • Title: I suspect a typo. We have $[\Bbb Q(i,\sqrt[4]{2}):\Bbb Q]=8$, see here. On the other hand, $\Bbb Q(i,\sqrt{-2})=\Bbb Q(i,\sqrt{2})$ and this has degree $4$, see above. – Dietrich Burde Jan 17 '22 at 17:40

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Extension $\Bbb Q (\alpha)$ over $\Bbb Q$ is indeed of degree 4, since $x^8-1 = (x^4-1)(x^4+1)$.

$x^4+1 = (x-\xi)(x-\xi^3)(x-\xi^5)(x-\xi^7)$ has as zeros the primitive 8th roots of unity, while $x^4-1=(x-1)(x+1)(x-i)(x+i)$ has as zeros only the 4th roots of unity with $i=\xi^2$.

Wuestenfux
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  • Why could the minimal polynomial not be $x^2+1$? Since $x^4-1$ factors further we could have that $\alpha$ satisfies either $x^2+1$ or $x^4+1$? How do we determine which one is the minimal polynomial, that is, which one it satisfies? – Mike Jan 17 '22 at 17:21
  • $x^4+1$ does not factor further over $\Bbb Q$. – Wuestenfux Jan 17 '22 at 17:26
  • But how do we know $\alpha$ satisfies $x^4+1$ and not $x^4-1$? – Mike Jan 17 '22 at 18:12