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I'm troubled by solving a homework problem:

If $\operatorname{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$ then $\mathbb{i} = \sqrt{-1} \notin K$

Any hints?

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    Think about that $\mathbb{Z}/4\mathbb{Z}$ has only a subgroup of order two, and about how one calculate the square root of a complez number. That you things combined will give you a contradiction when $i\in K$.2011-11-04
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    The point of the square root of a number is that is again a complex number, and that we have also its conjugate by $\sigma^2$. So we have its real and imaginary parts. Now, from there cannot you construct another cuadratic expansion of $\mathbb{Q}$ not contained in $\mathbb{Q}(i)$?2011-11-04
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    Perhaps we can sum up what Iasafro is saying: Such a Galois group implies there is only one quadratic subfield. If $i\in K$, can you construct two different quadratic subfields (thereby giving a contradiction)?2011-11-05
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    @Gurjott: What could $\sigma(i)$ be?2011-11-05

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