Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.