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Find the smallest normal extension (up to isomorphism) of $\mathbb Q(\sqrt[4]2)$ in $\overline {\mathbb Q}$ (the algebraic closure of $\mathbb Q)$

My atempt:

$\mathbb Q(\sqrt[4]{2},i)$ is a normal extension of $\mathbb Q(\sqrt[4]{2})$, because $\mathbb Q(\sqrt[4]{2},i)$ is a splitting field of $x^4-2\in \mathbb Q(\sqrt[4]{2})$.

Am I right? this is indeed the smallest normal extension of $\mathbb Q(\sqrt[4]{2})$? I found difficult to prove this is the smallest normal extension $\mathbb Q(\sqrt[4]{2})$.

I need a hand here


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up vote 3 down vote accepted

Well, $\,\Bbb Q(\sqrt [4]2\,,\,i)\,$ is algebraic, normal (as the irreducible $\,p(x)=x^4-2\,$ that has root in it in fact splits there) and, of course, separable, and $\,[\Bbb Q(\sqrt[4] 2\,,\,i):\Bbb Q]=4\,$ and, thus, it has the minimal non-trivial degree over the non-normal extension $\,\Bbb Q(\sqrt[4]2)\,$ , so yes: it is the minimal one in this sense.

Added: Of course, $\,\Bbb Q(\sqrt[4]2\,,\,i)/\Bbb Q(\sqrt[4] 2)\,$ is normal as well as its degree is two.

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I think you mean $[\mathbf Q(\sqrt[4]2, i):\mathbf Q(\sqrt[4]2)]=2$. – jathd Dec 4 '12 at 1:21
Of course, @jathd. Thanks – DonAntonio Dec 4 '12 at 2:50

But any quadratic extension of your field is normal, and the intersection of normal extensions is normal as well. So your given field $\mathbb Q(\sqrt2)$ is already the smallest normal extension of itself.

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I think you missed that little four in the root, @Lubin – DonAntonio Dec 4 '12 at 2:51
Doesn’t matter. The smallest normal extension of any field $k$ is still $k$. OP did not ask for the smallest normal extension of $\mathbb Q$ containing the given field, though it may be that that’s what he intended. – Lubin Dec 4 '12 at 3:31

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