A Cyclic Extension of degree 4 does not contain $i$

Question

The full question is: Let $K \subseteq \mathbb{C}$ be a cyclic exntesion of $\mathbb{Q}$ of degree 4. Prove that $i \notin K$.

I was thinking that since $K$ over $\mathbb{Q}$ is finite (has degree 4) and is separable (since it's Galois), by the Primitive Element Theorem, $K = \mathbb{Q}(\alpha)$ for some $\alpha \notin \mathbb{Q}$. So if I show that $i \neq \alpha$ then the statement holds. So suppose $i = \alpha$. Then $K = \mathbb{Q}(i)$. However, the degree of $\mathbb{Q}(i)$ over $\mathbb{Q}$ is not equal to 4. So $i \neq \alpha$.

Does this work? If not, please explain why/any ideas you may have. Thank you!

That plan won't work. For example $K=\Bbb{Q}(\sqrt2,i)=\Bbb{Q}(i+\sqrt2)$ has $\alpha\neq i$. $K$ is even Galois over $\Bbb{Q}$, but the Galois group isn't cyclic. — Jyrki Lahtonen, Aug 10 '17 at 19:34
But, this question has been asked here earlier. Gimme a minute. — Jyrki Lahtonen, Aug 10 '17 at 19:34
Complex conjugation fixes the unique quadratic subfield, which would be ${\mathbb Q}(i)$ if $K$ would contain $i$. — , Aug 10 '17 at 19:35
See here. Because I happened to answer that I should not cast the first vote to close this as a duplicate. — Jyrki Lahtonen, Aug 10 '17 at 19:38
I still suspect that I didn't find the first incarnation of this question :-( — Jyrki Lahtonen, Aug 10 '17 at 19:45

Tsemo Aristide · Answer 1 · 2017-08-10T20:46:13.880

You have extension $[K:\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]$. This extension is separable since the characteristic of $\mathbb{Q}$ is zero. Since $[\mathbb{Q}(i):\mathbb{Q}]=2$, you deduce that $K:\mathbb{Q}(i)$ is Galois. Let $a\in K$ such that $K=\mathbb{Q}(i)(a):\mathbb{Q}$ (primitive element theorem), and $f$ a generator of $Gal(K:\mathbb{Q})$, $P=(X-a)(X-f(a)(X-f^2(a))(X-f^3(a))$ is stable by $f$. This implies that $P\in\mathbb{Q}[X]$. The polynomial $P$ is irreducible since it is the minimal polynomial of $a$. You deduce that $K:\mathbb{Q}$ is Galois.

Let $h$ the generator of $Gal(\mathbb{Q}(i):\mathbb{Q})$, remark $h$ preserves $P$ and henceforth $K$. You have $f^2=h$. Since the order of $h$ is $2$ and there is a unique element of order $2$ in $\mathbb{Z}/4$. $f$ preserves $\mathbb{Q}(i):\mathbb{Q})$ since it preserves $X^2+1$, this impies that the restriction of $f^2$ to $\mathbb{Q}(i):\mathbb{Q})$ is the identity. Contradiction since $f^2=h$.

A Cyclic Extension of degree 4 does not contain $i$

1 Answers1