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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

1 Answers 1

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If $\mathrm{Gal}(K/\mathbb{Q}) \cong\mathbb{Z}_4$, then $[K:\mathbb{Q}]=4$ and $K$ has a unique subfield of degree $2$ over $\mathbb{Q}$. If $i\in K$, then this unique subfield must be $F=\mathbb{Q}(i)$.

Now, $K$ is a degree 2 extension of $F$, so there is an element $\beta$ of $F$ such that $K=F(\sqrt{\beta})$ (same argument as for quadratic extensions of $\mathbb{Q}$: you have an irreducible quadratic, the extension is given by the square root of the discriminant).

Now, $\beta = r + si$ for some $r,s\in\mathbb{Q}$. The square root of $\beta$ can be expressed as $p+qi$, where $$p = \frac{\sqrt{2}}{2}\sqrt{\sqrt{r^2+s^2} + r},\quad q = \frac{\pm\sqrt{2}}{2}\sqrt{\sqrt{r^2+s^2}-r}.$$ Complex conjugation is an automorphism of $K$, so we have both $\sqrt{\beta}$ and its complex conjugate in $K$. That means that we have both both $p$ and $q$ in $K$.

Now, look at $p^2$; it's in $K$. What else can you say?

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    Aren't these ideas about looking at square roots somewhat complex (excuse the pun)? If $i\in K$, we can simply say that the real subfield and $\mathbb{Q}(i)$ are two (distinct) quadratic subfields of $K$, contradicting the unique subgroup of index 2 in the Galois group.2011-11-05
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    @SteveD: Well.... yes; good point. )-: You should post it; I'll up vote and erase this one.2011-11-05
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    @SteveD - sorry to bother with this old post, but I can't figure what the other subfield is...there is $\mathbb{Q}(\sqrt{2})$ but what is the other ? (you wrote in the comment "the real subfield" I guess you don't mean $\mathbb{R}$ since the degree over $\mathbb{Q}$ is not 2...)2012-07-08
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    @Belgi: It means the intersection of $K$ with $\mathbb{R}$. Since $[K:\mathbb{Q}]=4$, but $K\not\subseteq \mathbb{R}$ (if we assume $i\in K$), then $K\cap\mathbb{R}$ is strictly larger than $\mathbb{Q}$, but strictly smaller than $K$. Hence $K\cap\mathbb{R}$ is a subfield of $K$ of degree $2$ over $\mathbb{Q}$; since it is a real field (contained in $\mathbb{R}$) it cannot equal $\mathbb{Q}(i)$, so $K$ contains both $\mathbb{Q}(i)$ and $K\cap\mathbb{R}$, contradicting that there is a *unique* subfield of degree $2$.2012-07-08
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    Do we know in the first place that $K\substack{\subset} \mathbb{C}$ ? I can't argue that $K\cap\mathbb{R}\neq\mathbb{Q}$...thanks for the help!2012-07-08
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    @Belgi: Yes, we can always assume that $K\subseteq\mathbb{C}$; because we can embed $K$ in the algebraic closure of $\mathbb{Q}$ that sits inside $\mathbb{C}$. As to why $K\cap \mathbb{R}\neq\mathbb{Q}$, if $K\cap\mathbb{R}=\mathbb{Q}$, then $K\subseteq \mathbb{Q}(i)$ (recall that $i\in K$ is an assumption), which would imply degree $2$.2012-07-08
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    @ArturoMagidin - I given this some thought today, Is it correct to say that: since the extension is finite it is algebric and since the algebric closure of $\mathbb{Q}$ must be contained in the he algebric closure of $\mathbb{R}=\mathbb{C}$ we can assume that $K$ is contained in $\mathbb{C}$ ? thank you for the help with this question (I had in on favourites to read later)2012-07-08
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    @Belgi: The algebraic closure is only unique *up to isomorphism*. However, any algebraic extension embeds into any particular algebraic closure. The collection of all elements of $\mathbb{C}$ that are algebraic over $\mathbb{Q}$ is algebraically closed (because $\mathbb{C}$ is algebraically closed), and algebraic over $\mathbb{Q}$, so it is *an* algebraic closure of $\mathbb{Q}$ (unique only up to isomorphism). Since $K$ is algebraic over $\mathbb{Q}$, it embeds into this algebraic closure of $\mathbb{Q}$, and so we may assume that it is contained in $\mathbb{C}$. (cont)2012-07-08
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    @Belgi: (cont) And once you think about it once, you never again go into all the detail about it, you just use the fact that you can consider any algebraic extension to be contained in $\mathbb{C}$.2012-07-08