Galois Basic Question (Cyclic Order 4 Extension of Q cannot contain i)

Question

Suppose $K$ is an extension of $\mathbb{Q}$ in $\mathbb{C}$, where $Gal(K/\mathbb{Q})$ is cyclic of order 4. Show that $i\notin K$.

($i$ is the imaginary number $i^2=-1$.)

My Galois theory is quite weak, hope someone can check if my attempt is correct.

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My attempt:

Suppose to the contrary $i\in K$. Let $\sigma\in Gal(K/\mathbb{Q})$.

Note that $\sigma (i)\sigma(i)=\sigma(i^2)=\sigma(-1)=-1$ since $\sigma$ fixes $\mathbb{Q}$.

This means that $\sigma(i)=i$ or $\sigma(i)=-i$. The first case is ruled out since $\sigma$ only fixes $\mathbb{Q}$.

So $\sigma(i)=-i$. This means that $\sigma(a+bi)=a-bi$, so $\sigma$ is effectively complex conjugation, which has order 2.

Since $\sigma$ was arbitrary, this contradicts that $Gal(K/\mathbb{Q})$ has an element of order 4.

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Is this ok?

Thanks.

Update: Now I see that my argument is clearly flawed. What would be the correct proof?

Answer 1

(fleshing this out now that OP worked out the details themself)

Because $K/\Bbb{Q}$ is Galois and complex conjugates share a minimal polynomial over $\Bbb{Q}$, we see that the complex conjugate of any element of $K$ is also an element of $K$. Therefore complex conjugation, call it $\sigma$, is an element of the Galois group $G=\operatorname{Gal}(K/\Bbb{Q})$.

Assume contrariwise that $i\in K$. Because $\sigma(i)=-i$ and $\sigma^2=1_G$, we can conclude that $\sigma$ is of order two. As $G$ is assumed to be cyclic, $H=\langle\sigma\rangle$ is the only subgroup of index two in $G$. By Galois correspondence the fixed field $M:=K^H=K\cap\Bbb{R}$ is the only quadratic intermediate field $\Bbb{Q}\subset M\subset K$.

But $\Bbb{Q}(i)$ would also be a quadratic intermediate field contradicting the above. Hence $i\notin K$.

Answer 2

This argument is not correct. All you've shown is that the restriction of $\sigma$ to $\mathbb{Q}(i)$ cannot have order $4$. But the case that $\sigma(i)=-i$ is (a priori) possible even if $\sigma$ has order $4$: in this case, you just would conclude that $\mathbb{Q}(i)$ is the fixed field of $\sigma^2$. This doesn't (obviously) mean that $\sigma^2$ is the identity on all of $K$.

Here is a sketch of a correct argument. We must have $K=\mathbb{Q}(i,\sqrt{a})$ for some $a\in\mathbb{Q}(i)$ that is not a square. In order for this $K$ to be Galois over $\mathbb{Q}$, $\sqrt{\overline{a}}$ must also be in $K$. Writing down explicitly what an element of $K$ looks like, you can show that this can only happen if $a=b^2c$ for some $b\in\mathbb{Q}(i)$ and some $c\in\mathbb{Q}$ (this is the hard step; at the moment I don't see a way to prove it without using unique factorization in $\mathbb{Z}[i]$). But then $K=\mathbb{Q}(i,\sqrt{c})$ can be shown to have Galois group $\mathbb{Z}/2\times\mathbb{Z}/2$, not $\mathbb{Z}/4$.

Answer 3

Here is a slick solution, I think. Consider generally a cyclic extension $F/k$ and try to embed it in an over-extension $K/F/k$ such that $K/k$ is cyclic. For simplification, suppose that $F/k$ has degree $p^n$, $K/F$ has degree $p$ and $k$ contains a primitive $p$-th root $\zeta$ of unity ($p$ a prime). Then, using Kummer theory, it can be shown that $K$ exists iff $\zeta$ is a norm in $F/k$ (see e.g. https://math.stackexchange.com/a/1691332/300700, where the situation is a bit more general). In our particular case here, $p=2$, $\zeta=-1$, $K/\mathbf Q$ is the given cyclic extension of degree 4. If $K$ contained $i$, take $F=\mathbf Q(i)$. Then $-1$ would be a norm in $F/\mathbf Q$, i.e. would be a sum of two squares in $\mathbf Q$ : impossible. Note that here, the previous embeddability criterion can be shown directly.