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?
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?
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?