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

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 $i\in K$. – Josué Tonelli-Cueto Nov 4 '11 at 23:13
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 $\mathbb{Q}(i)$? – Josué Tonelli-Cueto Nov 4 '11 at 23:39
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)? – user641 Nov 5 '11 at 1:32
@Gurjott: What could $\sigma(i)$ be? – Kevin Nov 5 '11 at 4:36

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. – user641 Nov 5 '11 at 4:49
@SteveD: Well.... yes; good point. )-: You should post it; I'll up vote and erase this one. – Arturo Magidin Nov 5 '11 at 4:51
@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...) – Belgi Jul 8 '12 at 0:57
@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 $2$. – Arturo Magidin Jul 8 '12 at 1:13
@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 $2$. – Arturo Magidin Jul 8 '12 at 17:44