# Norm -1 in the extension $\,E[i]/E\,$ , where $\,E=\Bbb Q(\zeta)\,,\,\, \zeta^5 = 1$

Denote by $\zeta = \exp(2\pi i/5)$ the primitive root of unit of order 5 ($\zeta^5=1, \zeta \ne 1$). Let $E = \mathbb{Q}[\zeta]$. Then $i = \sqrt{-1} \notin E$. Let $L = E[i]$. We want to show that $-1$ is a norm from $L$ to $E$.

There is an hint to use $( \zeta + \zeta^4 ) (1 + \zeta^2) = \zeta + \zeta^3 + \zeta^4 + \zeta$ but we don't understand how it helps.

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What do you mean a number is a norm? – DonAntonio Feb 5 '13 at 12:55
That there is $\alpha \in L$ such that $$-1 = N_{L/E}(\alpha)$$ – LinAlgMan Feb 5 '13 at 13:08
Oh, I see: you meant $\,-1\,$ is the norm of some element in $\,L\,$ over $\,E\,$... – DonAntonio Feb 5 '13 at 13:11

I think I have a better hint : $(\zeta^6 + \zeta^2)(1+ \zeta^2) = \zeta + \zeta^2 + \zeta^3 + \zeta^4 = -1$.
We tried to link this to the norm. Here the basis of the extension $L/E$ is $\{ 1 , i \}$. Can you give more details? – LinAlgMan Feb 5 '13 at 15:12
@LinAlgMan : $N(a+ib) = a^2 + b^2$ (norms from $L$ are sums of two squares of elements of $E$); and $N(xy) = N(x)N(y)$ (a product of two norms from $L$ is still a norm from $L$). Now the equality shows how $-1$ is a product of two sums of two squares of elements of $E$, hence it's a norm. – mercio Feb 5 '13 at 15:24
Thanks, the thing that confused us was that we took $\sigma \in \mathrm{Gal}(L/E)$ as the usual conjugation (with $\zeta \in \mathbb{C}$) and not just $\sigma(i) = -i$ and $\sigma(\zeta)=\zeta$ since $\zeta$ is in the fixed field $E$. – LinAlgMan Feb 6 '13 at 7:54