I trying to solve a problem which goes like this:
Let $F$ be a field that does not contain a primitive 4th root of unity and let $L=F(\sqrt{a})$, for some $a\in F^*\setminus F^{*2}$. Then the following are equivalent:
(a) $a$ is a sum of two squares.
(b) $-1=N_{L/F}(\alpha)$, for some $\alpha\in L$.
(c) $a=N_{L/F}(\alpha')$, for some $\alpha'\in L$.
(d) $N_{L/F}(b)\equiv a \text{ mod } F^{*2}$, for some $b\in L$.
(e) $K/F$ is a cyclic extension of degree 4, where $K=L(\sqrt{b})$ ($b$ from (d) above)
(f) $L$ lies in a cyclic extension of $F$ of degree 4.
I am stuck on the implication $(f)\implies (a)$. This is what I have so far:
Suppose $K$ be a cyclic extension of $F$ of degree 4 containing $L$. Then $[K:L]=2$, i.e., $\exists~b\in L^*\setminus L^{*2}$ such that $K=L(\sqrt{b})$. It can be easily shown that $b\not\in F$. There are elements $\gamma,\delta\in F$ such that $b=\gamma+\delta\sqrt{a}$. The minimal polynomial of $\sqrt{b}$ over $F$ is $x^4-2\gamma x+\gamma^2-\delta^2a$ and has the roots $\pm \alpha,\pm \beta$, where $\alpha=\sqrt{\gamma+\delta\sqrt{a}}$ and $\beta=\sqrt{\gamma-\delta\sqrt{a}}$. Since $K/F$ is cyclic (and, in particular, normal) $\gamma^2-\delta^2a=\alpha^2\beta^2\in K^{*2}$, i.e., $\sqrt{\gamma^2-\delta^2a}\in K$.
I am unsure about what to do next. It would be very helpful if anyone could give me some hint.
Given $K/L/F$ with $K/F$ cyclic of degree $4$ and $L=F(\sqrt{a})=F(1/\sqrt{a}),a\in F$,
I'd say that $K=L(\sqrt{u+v/\sqrt{a}})$,
$\sqrt{u+v/\sqrt{a}}\sqrt{u-v/\sqrt{a}}=\sqrt{u^2-v^2/a}$ lies in the unique quadratic subfield, $L$,
So $\sqrt{u^2-v^2/a}=r+s/\sqrt{a}$ with $r,s\in F$.
$-\sqrt{u^2-v^2/a}$ is $F$-conjugate to $\sqrt{u^2-v^2/a}$ so $r=0$.
And hence $$u^2-v^2/a = s^2/a, \qquad a =(v/u)^2 + (s/u)^2$$