Let $E/F$ be a Galois extension of fields with Galois group $G=\{1,\sigma_1,\sigma_2,\sigma_3\}\simeq (\mathbb{Z}/2\mathbb{Z})^2$. Let $F_i=E^{\sigma_i}$ and $N_i=N_{F_i/F}(F_i^{\times})$. Show that $N_1N_2N_3=\{x\in F^{\times}| x^2\in N\}$, where $N=N_{E/F}(E^{\times})$.
I have no idea about this problem. Any hints are appreciated. Thanks in advance.
The previous answer was very, very wrong, but I finally managed to find a fix. The old answer is at the end for legacy, but the valid proof is as follows:
First, by the identity in Jyrki Lahtonen’s comment, $(N_1N_2N_3)^2 \subset N$, entailing $LHS \subset RHS$.
Now, let $x \in F^{\times}$ be such that $N_{E/F}(z)=x^2$ for some nonzero $z \in E$. Let $1 \leq i \leq 3$, let $j \neq i$ be between $1,2,3$, and note that $\frac{N_{E/F_i}(z)}{x} \in F_i$ has norm $1$ over $F$.
As $F_i/F$ is quadratic with only nonzero automorphism $\sigma_j$, by ”Hilbert 90” there exists $y_i \in F_i$ such that $\frac{z\sigma_i(z)}{x}=\frac{y_i}{\sigma_j(y_i)}$.
By taking $i=1,3$, $j=2$ in both equations, and taking the quotient of the two equations, it follows $\frac{\sigma_1(z)}{\sigma_2\circ \sigma_1(z)}=\frac{(y_1/y_3)}{\sigma_2(y_1/y_3)}$, which entails $\frac{y_1}{y_3}\in F_2 \sigma_1(z)$.
Now, by taking cyclic product, it follows $1=\frac{y_1}{y_3}\frac{y_3}{y_2}\frac{y_2}{y_1} \in F_1F_2F_3\sigma_1(z)\sigma_2(z)\sigma_3(z)=F_1F_2F_3\frac{x^2}z$, and therefore $z \in F_1F_2F_3$.
By the identity in Jyrki Lahtonen’s comment, it follows $x^2 \in (N_1N_2N_3)^2$, hence $x$ or $-x$ is in $N_1N_2N_3$. To conclude, we only need to show that $-1 \in N_1N_2N_3$, as stated in Jyrki Lahtonen’s comment.
We treat the case where the characteristic is not $2$ (it’s obvious otherwise).
Let $j \neq i$. $\sigma_j$ is a nontrivial $F$-linear symmetry of $F_i$ so has an eigenvector for $-1$, which we denote as $u_i$. Note that $u_i$ is also a $-1$-eigenvector for the last element of $G$ ( which acts as $\sigma_j$). Therefore $\sigma_3(u_1u_2)=(-u_1)(-u_2)=u_1u_2$, and $\sigma_1(u_1u_2)=u_1(-u_2)=-u_1u_2$, so we can take $u_3=\frac{1}{u_1u_2}$.
Now, $N_{F_1/F}(u_1)N_{F_2/F}(u_2)N_{F_3/F}(u_3)=(-u_1^2)(-u_2^2)(-u_3^2)=-1$ and we are done.
Old answer:
Write $$y=\left(\frac{\sigma_2(y)}{\sigma_1(y)}\right)(y\sigma_1(y))(\sigma_1(y)\sigma_3(y))\left(\frac{1}{\sigma_1(y)\sigma_2(y)}\right).$$
It follows that if $K =\{x \in E^{\times},\, N_{E/F}(x)=1\}$, $E=KF_1^{\times}F_2^{\times}F_3^{\times}$.
As a consequence, for a given $x \in F^{\times}$, $x \in N$ iff $x \in N_{E/F}(F_1^{\times}F_2^{\times}F_3^{\times})$, iff (by the identity in Jyrki Lahtonen’s comment) $x \in (N_1N_2N_3)^2$.
It remains to show, to conclude, that, as stated in Jyrki Lahtonen’s comment, $-1 \in N_1N_2N_3$. We treat the case where the characteristic is not $2$ (it’s obvious otherwise).
Let $j \neq i$. $\sigma_j$ is a nontrivial $F$-linear symmetry of $F_i$ so has an eigenvector for $-1$, which we denote as $u_i$. Note that $u_i$ is also a $-1$-eigenvector for the last element of $G$ ( which acts as $\sigma_j$). Therefore $\sigma_3(u_1u_2)=(-u_1)(-u_2)=u_1u_2$, and $\sigma_1(u_1u_2)=u_1(-u_2)=-u_1u_2$, so we can take $u_3=\frac{1}{u_1u_2}$.
Now, $N_{F_1/F}(u_1)N_{F_2/F}(u_2)N_{F_3/F}(u_3)=(-u_1^2)(-u_2^2)(-u_3^2)=-1$ and we are done.