Hilberts Theorem (norm group) The theorem says the following:  The map $N$ is a group homomorphisim from the multiplicative group of $\mathbb{Q}^{x}[i]$ to the multiplicative group of $\mathbb{Q}^{x}$ and has kernel $\lbrace \dfrac{z}{\sigma(z)} | z \in \mathbb{Q}^{x}[i] \rbrace $
I understand the proof but there is one part of the proof I do not understand and that is showing that given a element in the kernel of $N$ we show that it must be in the form of $\dfrac{z}{\sigma(z)}$. Is there some way to show this?
$z=x+iy$
$\sigma(z) = x-iy$
$n(z)=z \sigma(z)$
 A: This is a special case of Hilbert's Theorem 90. Because you are just looking at this special case, there is a very fun way to see this.
If you plot points in $\mathbb{Q}(i)$ in the complex plane, saying that a point is in the kernel of the norm map means precisely that it is a point with rational coordinates on the unit circle.  There is a venerable old result describing exactly such points -- the rational parameterization of the unit circle.  It says that all such points have coordinates of the form
$$\left( \frac{m^2-n^2}{m^2+n^2}, \frac{2mn}{m^2+n^2} \right)$$
for some integers $m$ and $n$.  There are many easily googlable sources for this, for instance:
http://math.rice.edu/~evanmb/math499spring10/vigrenotes3.pdf
Believing this, you want to show that such a point must have the form $\tfrac{z}{\sigma(z)}$ for some $z$.  If you try $z=m+in$, where $m$ and $n$ are integers giving the coordinates of your element of the kernel above, then
$$
\frac{z}{\sigma(z)} = \frac{m+in}{m-in} = \frac{(m+in)^2}{(m-in)(m+in)} = \frac{m^2-n^2}{m^2+n^2} + i \frac{2mn}{m^2+n^2}
$$ 
This point has the same coordinates as your point in the kernel, so you've found your $z$.
P.S. Noam Elkies dicusses this here
http://www.math.harvard.edu/~elkies/Misc/hilbert.pdf
but takes the point of view of establishing the above parameterization of the unit circle from Hilbert's Theorem.  
