# Thread: Show x is a unit iff N(x)=1 in Z[w]

1. ## Show x is a unit iff N(x)=1 in Z[w]

Let $\displaystyle w = \frac {-1}{2} + \frac {3^{1/2}}{2}i$, let $\displaystyle N:Z[w] \rightarrow Z \ by\ N(a+bw) = a^2 - ab + b^2$ and suppose that x is in Z[w]. Show that x is a unit iff N(x)=1.

My proof so far:

Assume xy = 1 for y in Z[w].

1 = N(1) = N(xy) = N(x)N(y), implies that N(x) = N(y) = 1.

Conversely, let x = a + bw with N(x) = 1.

Then N(a+bw) = 1 = a^2 - ab + b^2

Then that should equals to (a+bw)(???)

Any hints? Thanks.

Let $\displaystyle w = \frac {-1}{2} + \frac {3^{1/2}}{2}i$, let $\displaystyle N:Z[w] \rightarrow Z \ by\ N(a+bw) = a^2 - ab + b^2$ and suppose that x is in Z[w]. Show that x is a unit iff N(x)=1.

My proof so far:

Assume xy = 1 for y in Z[w].

1 = N(1) = N(xy) = N(x)N(y), implies that N(x) = N(y) = 1.

Conversely, let x = a + bw with N(x) = 1.

Then N(a+bw) = 1 = a^2 - ab + b^2

Then that should equals to (a+bw)(???)

Any hints? Thanks.
You're looking for an element (???) such that (a+bw)(???) = 1, where $\displaystyle \textstyle w=-\frac12+\frac{\sqrt3}2i$. The standard way of doing this is to find the inverse of a+bw by rationalising the denominator:

$\displaystyle \frac1{a+bw} = \frac{a+b\bar{w}}{(a+bw)(a+b\bar{w})}$, where $\displaystyle \bar{w}$ is the complex conjugate of w. When you multiply out the expression in the denominator, you'll discover the reason for defining N(a+bw) in that way.

3. Assuming that $\displaystyle N$ is a euclidean norm, then here is my proof..

If $\displaystyle x$ is a unit, then $\displaystyle xx^{-1} = 1$. So,
$\displaystyle 1 = N(1) = N(xx^{-1}) = N(x)N(x^{-1})$. Since, for any $\displaystyle x$, $\displaystyle N(x) \geq 0$, we have $\displaystyle N(x) = 1$.

Coversely, suppose $\displaystyle N(x) = 1$. Let $\displaystyle x = a + bw$, where $\displaystyle a,b \in \mathbb{Z}$. Now, $\displaystyle N(x) = N(a + bw) = a^2 - ab + b^2 = 1$
This means that $\displaystyle (a,b)$ forms a solution set
$\displaystyle \{ (\pm 1,0), (0, \pm 1), (-1,-1) \}$.

For $\displaystyle (a,b) = (\pm 1,0) \implies x = \pm 1$ which are units.
For $\displaystyle (a,b) = (0, \pm 1) \implies x = \pm w = \frac{\mp 1}{2} + \frac{ \pm \sqrt 3}{2}i$ which are also units.
For $\displaystyle (a,b) = (\pm 1,\pm 1) \implies x = \pm 1 \pm 1w = -1 + \frac{1}{2} - \frac{\sqrt 3}{2}i = \frac{-1}{2} - \frac{\sqrt 3}{2}i$ which is also a unit.

In all cases, x is a unit. QED.

4. I was working with Opalg's proof, and now I got to this point:

$\displaystyle 1=a^2-ab+b^2=(a+bw)( \frac {a+b \bar {w}}{a^2-ab+b^2} )$

But ain't I looking for (a+bw)(u+vw) for some integers u and v? And how do I turn that into w? Thank you!

$\displaystyle 1=a^2-ab+b^2=(a+bw)( \frac {a+b \bar {w}}{a^2-ab+b^2} )$
Don't forget that you're assuming that N(a+bw)=1. So $\displaystyle (a+bw)(a+b \bar {w}) = 1$. You also know that $\displaystyle \textstyle\bar{w} = -\frac12-\frac{\sqrt3}2i = -1-w$. Therefore $\displaystyle (a+bw)(a-b-bw) = 1$, and you can take u=a–b and v=–b.