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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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    @Gurjott: What could $\sigma(i)$ be?2011-11-05

1 Answers 1

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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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    @Belgi: (cont) And once you think about it once, you never again go into all the detail about it, you just use the fact that you can consider any algebraic extension to be contained in $\mathbb{C}$.2012-07-08