number theory - Computing $mathcal{O}_K*$ by looking at the subfields

I have a question regarding some notes that my lecturer made on Algebraic Number Theory. We want to compute the unit group $\mathcal{O}_{K}*$ for the field $K=\mathbb{Q}(\sqrt{-2},\sqrt{3})$. We have that the rings of integers of the subfields are $\mathbb{Z}[\sqrt{-2}]$, $\mathbb{Z}[\sqrt{3}]$ and $\mathbb{Z}[\sqrt{6}]$ which have groups of units respectively $<-1>$, $<-1>$ and $<-1,2+\sqrt{3}>$. Then there is written:
'Let $\nu \in \mathcal{O}_K^*$. Taking the norms to the three quadratic subfields, we get:'
\begin{align}
\nu^2 \in <-1,2+\sqrt{3}> \quad (*)
\end{align}
The steps that follow to compute $\mathcal{O}_K*$ I understand, but this thing above I don't get.
If we take the norms of $\nu$ to the subfields we would get:

\begin{align*}
N_{\mathbb{Q}(\sqrt{-2})}^K(\nu)=\nu\sigma_2(\nu) \\
N_{\mathbb{Q}(\sqrt{3})}^K(\nu)=\nu\sigma_3(\nu) \\
N_{\mathbb{Q}(\sqrt{-6})}^K(\nu)=\nu\sigma_4(\nu) \\
\end{align*}
where $\sigma_2$, $\sigma_3$ and $\sigma_4$ are the automorphisms that fix the respective subfields. Since we know that relative norms of units are also units, we can conclude that:
\begin{align*}
\nu\sigma_2(\nu) \in <-1> \\
\nu\sigma_3(\nu) \in <-1, 2+\sqrt{3}>\\
\nu\sigma_4(\nu) \in <-1>

\end{align*} But I don't see how to use this to come to (*).

Answer

I write $K=\mathbb Q(\sqrt{-2},\sqrt{3})$, $E=\mathbb Q(\sqrt{-2})$, $F=\mathbb Q(\sqrt{-3})$, $G=\mathbb Q(\sqrt{-6})$.

Then $$N^K_E(\nu)\cdot N^K_G(\nu)=\dfrac{N^K_{\mathbb Q}(\nu)\cdot\nu^2}{N^K_F(\nu)}=\pm 1$$

Also $N^K_{\mathbb Q}(\nu)=\pm1$ and $N^K_F(\nu)\in\langle-1,\varepsilon\rangle$. (Note: the norms here can't actually be negative, but it doesn't really matter.)

So rearrange to get what you want.