I need to show that $\mathbb{Q}(2^{1/4}) $ isomorphic to $\mathbb{Q}(2^{1/4}i) $. My issue is showing $2^{1/4}i $ is in $\mathbb{Q}(2^{1/4}) $. I dont see how there could be an imaginary part in $\mathbb{Q}(2^{1/4}) $.
$\mathbb{Q}(2^{1/4}) $ isomorphic to $\mathbb{Q}(2^{1/4}i) $?
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galois-theory
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1$Q(r)$ indeed doesn't comtain imaginary numbers for a real $r$. "Isomorphic" doesn't mean "equal". – 2017-02-06
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0Let $F,L$ and $K$ three field. $L$ and $F$ are isomorphic $\iff$ $L/K$ and $F/K$ has same degree extension (we suppose that $K$ is a subfield of $F$ and $L$). Here $[\mathbb Q(2^{1/4}):\mathbb Q]=[\mathbb Q(i2^{1/4}):\mathbb Q]=4$ and thus, the claim follow. – 2017-02-06
1 Answers
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Showing that two subfields of $ \mathbb C $ are isomorphic is not the same thing as showing that they are equal. Indeed, $ \mathbb Q(2^{1/4}) \neq \mathbb Q(2^{1/4} i) $ as subfields of $ \mathbb C $, but they are nevertheless isomorphic.
Hint: Use the result that if $ f $ is the minimal polynomial of $ a $ over some field $ F $, where $ a $ lies in a field extension $ E/F $, then $ F[x]/(f) \cong F(a) $.