let $k$ be a field of characteristic $\neq 2$. Let $c\in k, c\notin k^2$. Let $F=k(\sqrt{c})$ . Let $\alpha=a+b\sqrt{c}$ with $a,b\in k$ not both $a,b=0$. Let $E=F(\sqrt{\alpha})$. Prove that the following are equivalent :
- $E$ is Galois over $k$.
- $E=F(\sqrt{\alpha'})$ where $\alpha'=a-b\sqrt{c}$.
- Either $\alpha\alpha'\in k^2$ or $c\alpha \alpha'\in k^2$.
Show that when these conditions are satisfied, then $E$ is cyclic over $k$ of degree $4$ iff $c\alpha\alpha'\in k^2$.
What i have tried so far is :
$E=F(\sqrt{a+b\sqrt{c}})=k(\sqrt{c},\sqrt{a+b\sqrt{c}})=k(\sqrt{a+b\sqrt{c}})$
As $\sqrt{a-b\sqrt{c}}$ is another root of minimal polynomial of $\sqrt{a+b\sqrt{c}}$ and as $E/k$ is galois we see that $E=F(\sqrt{a-b\sqrt{c}})$ (adjoining any root should give same extension)..
So, first statement implies second..
Suppose $\alpha\alpha'\in k^2$ i.e., $a^2-b^2c\in k^2$ then i can write $\sqrt{a+b\sqrt{c}}=\sqrt{m}+\sqrt{n}$ with $m,n\in k$.
To be precise, $m=\frac{1}{2}(a+\sqrt{a^2-b^2c})$ and $n=\frac{1}{2}(a-\sqrt{a^2-b^2c})$
Then i have $E=k(\sqrt{a+b\sqrt{c}})=k(\sqrt{m}+\sqrt{n})=k(\sqrt{m},\sqrt{n})$
We have a result that says $k(\sqrt{m},\sqrt{n})$ is of degree $4$ if $mn $ is not a square in $k$..
In this case we have $mn=c$ which is not a square in $k$ so $E$ is of degree $4$ over $k$... More over i can see in this case we have galois group to be $\mathbb{Z}_2\times \mathbb{Z}_2$.
So, i have proved that if $c\alpha\alpha'\notin k^2$ then Galois group is not cyclic.. As there are only two possible groups of order $4$ i have prove that $E$ is cyclic over $k$ of degree $4$ iff $c\alpha\alpha'\in k^2$.
I have no idea how do you proceed proving second statement implies the third and third implies first..
Please give some hints..
You've shown $(1 \Longrightarrow 2)$ so I will only discuss the rest.
Hints:
For the implication $(2 \Longrightarrow 3)$ show the product $\sqrt{\alpha}\sqrt{\alpha'}\in E$ is actually in $F$. Consider how the non-trivial automorphism of $F/k$ acts on $\sqrt{\alpha}\sqrt{\alpha'}$, and show either $\sqrt{\alpha}\sqrt{\alpha'}\in k$ or $\sqrt{c}\sqrt{\alpha}\sqrt{\alpha'}\in k$.
For the implication $(2 \Longrightarrow 1)$ investigate the possibilities for $\sigma(E)$, for $\sigma \in Gal(\overline{E}/k)$, where $\overline{E}$ is some fixed algebraic closure of $E$. What are the possibilities for $\sigma(\sqrt{\alpha})$? Recall that $E/k$ is Galois if and only if $\sigma(E)=E$ for all such $\sigma$.
For the implication $(3 \Longrightarrow 2)$ notice that $F(\sqrt{\alpha})=F(\sqrt{\alpha'})$ if $\sqrt{\alpha} \sqrt{\alpha'} \in F$, or equivalently, if $\alpha \alpha' \in F^2$.
When these conditions are satisfied, it's possible to given an automorphism of $E$ by specifying how it acts on $\sqrt{c}$ and on $\sqrt{\alpha}$. One can show $\sigma$ must have order $2$ if it fixes $\sqrt{\alpha}\sqrt{\alpha'}$.
Solutions:
$(2 \Longrightarrow 3):$The non-trivial automorphism of $E/F=F(\sqrt{\alpha})/F$ sends $\sqrt{\alpha}$ to $-\sqrt{\alpha}$. If $E=F(\sqrt{\alpha'})$, it also sends $\sqrt{\alpha'}$ to $-\sqrt{\alpha'}$. Therefore it fixes $\beta = \sqrt{\alpha}.\sqrt{\alpha'}$ and so $\beta \in F$. The non-trivial automorphism of $F/k$ sends $\sqrt{c}$ to $-\sqrt{c}$. Thus it sends $a+b\sqrt{c}$ to $a-b\sqrt{c}$, so it fixes $\beta^2=\alpha \alpha'$, and therefore it either fixes $\beta$ or sends it to $-\beta$. If it fixes $\beta$, $\alpha \alpha'=\beta^2$ is a square in $k$. Otherwise it sends $\beta$ to $-\beta$, so it fixes $\sqrt{c}\beta$, and in that case $c\alpha \alpha'=(\sqrt{c} \beta)^2$ is a square in $k$.
$(2 \Longrightarrow 1):$ Suppose $\sigma$ is an automorphism of some algebraic closure $\overline{E}$ of $E$ that fixes $k$. We have $\sigma(\sqrt{c})=\pm \sqrt{c}$. Either $\sigma(\sqrt{c})=\sqrt{c}$, in which case $\sigma(\sqrt{\alpha})=\pm \sqrt{\alpha}$, or $\sigma(\sqrt{c})=-\sqrt{c}$, in which case $\sigma(\sqrt{\alpha})=\pm \sqrt{\alpha'}$. Thus either $\sigma(F(\sqrt{\alpha}))=F(\sqrt{\alpha})$ or $\sigma(F(\sqrt{\alpha}))=F(\sqrt{\alpha'})$. Since the latter case does occur for some $\sigma$, it follows that $F(\sqrt{\alpha})=F(\sqrt{\alpha'})$ if and only if $\sigma(F(\sqrt{\alpha}))=F(\sqrt{\alpha})$ for every $\sigma$, i.e. if and only if $E$ is Galois over $k$.
$(3 \Longrightarrow 2):$ If either $\alpha \alpha'$ or $c \alpha \alpha'$ belong to $k^2$, then $\alpha \alpha' \in F^2$, so that $\beta=\sqrt{\alpha}\sqrt{\alpha'} \in F$. This implies $\sqrt{\alpha'}=\beta/\sqrt{\alpha} \in E$. Since $\sqrt{\alpha'}\in E$ has degree $2$ over $F$, $E=F(\sqrt{\alpha'})$.
Suppose these conditions are satisfied and $Gal(E/k)$ is cyclic of order $4$, generated by $\sigma$. Note that $\sigma$ is determined by its action on $\sqrt{c}$ and $\sqrt{\alpha}$. Now $\sigma(\sqrt{c})=-\sqrt{c}$ necessarily otherwise all powers of $\sigma$ act trivially on $F$, so $\sigma(\sqrt{\alpha})=\pm \sqrt{\alpha'}$. If $\alpha \alpha' \in k^2$, then $\sigma$ fixes $\sqrt{\alpha}\sqrt{\alpha'}$. In that case it would follow that $\sigma^2$ fixes $\sqrt{\alpha}$ and $\sigma$ would have order $2$, resulting in a contradiction. Therefore $c\alpha \alpha' \in k^2$ by the third property above. Conversely, if $c \alpha \alpha' \in k^2$, then we have an automorphism of $E/k$ of order $4$, determined by $\sigma(\sqrt{c})=-\sqrt{c}$ and $\sigma(\sqrt{\alpha})=-\sqrt{\alpha'}$.