Eisenstein Ring's Norm

Theorem

Let’s consider a function on the Eisenstein ring $\mathbb{Z}[ \omega ]$.

[1]: If defined as $N(x + \omega y) := x^2 - xy + y^2$, then $N$ becomes the multiplicative norm of $\mathbb{Z}[ \omega ]$.
[2]: $\mathbb{Z}[ \omega ]$ is a Euclidean domain.
[3]: The only unit of $\mathbb{Z}[ \omega ]$ is $\pm 1, \pm \omega, \pm \omega^2 $.

Description

The study of Eisenstein integers can be greatly facilitated with the help of [abstract algebra](../../categories/abstract algebra). Proving [2] with the norm $N$ defined in an integral domain makes it almost equivalent to proving the fundamental theorem of arithmetic extended to Eisenstein primes since ED is UFD. This is similar to algebraically explaining the fundamental theorem of arithmetic as $\mathbb{Z}$ being UFD.

Proof

[1]

Definition of multiplicative norm:
(i): $N (\alpha) = 0 \iff \alpha = 0$
(ii): $N ( \alpha \beta ) = N ( \alpha ) N ( \beta )$

We need to show that $N$ becomes a multiplicative norm and that $\mathbb{Z}[ \omega ]$ actually becomes an ID. There is no need to first prove that it’s an ID as the theorem requiring a norm to be defined isn’t needed yet. Let’s say $a,b,c,d \in \mathbb{Z}$ with respect to $\alpha := a + \omega b$, $\beta := c + \omega d$.

Part (i). $N (\alpha) = 0 \iff \alpha = 0$

If we set $a := (A+B)$, $b:= (A-B)$, then $$ \begin{align*} & N(\alpha) = a^2 - ab+ b^2 = 0 \\ \iff & (A+B)^2 - (A+B)(A-B) + (A-B)^2 = 0 \\ \iff & A^2 + 3 B^2 = 0 \\ \iff & A = B = 0 \\ \iff & a = b = 0 \\ \iff & \alpha = a + \omega b = 0 \end{align*} $$

Part (ii). $N ( \alpha \beta ) = N ( \alpha ) N ( \beta )$

Since the product of Eisenstein integers is calculated as $( a + \omega b )( c + \omega d) = (ac - bd) + \omega (ad - bd + bc)$, $$ \begin{align*} N ( \alpha \beta ) =& N \left( ac - bd + \omega (ad -bd + bc) \right) \\ =& (ac - bd)^2 - (ac -bd)(ad - bd + bc) + (ad - bd + bc)^2 \\ =& (ac -bd)( ac -ad - bc) + (ad - bd + bc)^2 \\ =& (ac -bd)( ac -K ) + (K - bd )^2 \\ =& a^2 c^2 - acbd - acK + bdK + k^2 - 2Kbd + bd^2 \\ =& a^2 c^2 - acbd - acK - bdK + k^2 + bd^2 \\ =& a^2 c^2 - acbd - (ac + bd)(ad + bc) + (ad + bc)^2 + bd^2 \\ =& a^2 c^2 - acbd -a^2 c d - a b c^2 - abd^2 - b^2 c d +a^2 d^2 + 2 adbc + b^2 c^2 + bd^2 \\ =& a^2 c^2 - a^2 c d + a d^2 - abc^2 + abcd - abd^2 + b^2 c^2 - b^2 c d + b^2 d^2 \\ =& (a^2 - ab + b^2)(c^2 - cd + d^2) \\ =& N ( \alpha ) N ( \beta ) \end{align*} $$ where $K := (ad + bc)$ is.

Part 3. $Z[ \omega ]$ is an ID

Assume $\alpha$, $\beta$ are not $0 \in \mathbb{Z}[ \omega ]$ yet $\alpha \beta = 0$. Then, according to Part (i)~(ii), $$ N(\alpha) N(\beta) = N(\alpha \beta ) = N(0) = 0 $$ As $N(\alpha)$, $N(\beta)$ are elements of the integral domain $\mathbb{Z}$, to meet $N(\alpha) N(\beta) = 0$, either $N(\alpha)$ or $N(\beta) $ must be $0$, which contradicts the assumption, thus $Z[ \omega ]$ is also an integral domain.

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[2]

Definition of Euclidean norm:
(i): For all $a,b \in D (b \ne 0 )$, there exist $q$ and $r$ such that $a = bq + r$ is satisfied. In this case, $r = 0$ or $\nu (r) < \nu (b)$ must be met.
(ii): For all $a,b \in D (b \ne 0 )$, $\nu ( a ) \le \nu ( ab )$

Let’s say $\nu := N$ and show that $N$ becomes the Euclidean norm of $\mathbb{Z}[ \omega ]$. Assume $\beta \ne 0$.

Part (i). $\exists \sigma, \rho : \alpha = \beta \sigma + \rho$

For some $r, s \in \mathbb{Q}$, let’s say $\displaystyle {{ \alpha } \over { \beta }} = r + \omega s$. For $r, s$ close to an integer $q_{1} , q_{2} \in \mathbb{Z}$, let’s consider $\sigma$ and $\rho$ as follows. $$ \sigma := q_{1} + \omega q_{2} \\ \rho := \alpha - \beta \sigma $$

Case 1. $\rho = 0$
Since $\alpha = \beta \sigma$, there’s no more to show. We found $\displaystyle \sigma = {{ \alpha } \over { \beta }} = q_{1} + \omega q_{2}$ and $\rho = 0$.
Case 2. $N ( \rho ) < N ( \beta )$
By definition of $q_{1}$ and $q_{2}$, $$ | r - q_{1} | \le {{1} \over {2}} \\ | s - q_{2} | \le {{1} \over {2}} $$ thus, $$ \begin{align*} N \left( {{ \alpha } \over { \beta }} - \sigma \right) =& N \left[ (r + \omega s) - (q_{1} + \omega q_{2}) \right] \\ =& N \left[ (r - q_{1} ) + \omega ( s - q_{2}) \right] \\ =& (r - q_{1})^2 - (r - q_{1}) (s - q_{2}) + (s - q_{2})^2 \\ \le & |r - q_{1}|^2 + | r - q_{1} | | s - q_{2} | + | s - q_{2} |^2 \\ \le & \left( {{1} \over {2}} \right)^2 + \left( {{1} \over {2}} \right) \left( {{1} \over {2}} \right) + \left( {{1} \over {2}} \right)^2 \\ =& {{3} \over {4}} \end{align*} $$ Accordingly, $$ \begin{align*} N ( \rho ) =& N(\alpha - \beta \sigma ) \\ =& N \left( \beta \left( {{ \alpha } \over { \beta }} - \sigma \right) \right) \\ =& N (\beta) N \left( {{ \alpha } \over { \beta }} - \sigma \right) \\ \le & N(\beta) {{3} \over {4}} \\ \le & N (\beta) \end{align*} $$ ensuring that $\rho$, $\sigma$ exist satisfying either $\rho = 0$ or $N(\rho) \le N (\beta)$.

Part (ii). $N ( \alpha ) \ge N ( \alpha ) N ( \beta )$

Since $\beta \ne 0 \implies N ( \beta ) \ge 1$, $$ \begin{align*} N ( \alpha ) \le & N ( \alpha) \cdot 1 \\ \le & N(\alpha) N(\beta) \\ =& N ( \alpha \beta ) \end{align*} $$

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[3]

$20190907\_230034.png$

According to the properties of multiplicative norm, since $N(1) = 1$ and if $u \in \mathbb{Z}[ \omega ]$ is a unit, then $| N ( u ) | = 1$, by contraposition, if not $| N(u) | = 1$, then $u$ is not a unit. The case satisfying $u := x + \omega y$ for $N(u) = x^2 - xy + y^2 = 1$ is $$ \begin{align*} e^{0 \pi i / 3 } =& 1 \\ e^{1 \pi i / 3 } =& - \omega^2 \\ e^{2 \pi i / 3 } =& \omega \\ e^{3 \pi i / 3 } =& -1 \\ e^{4 \pi i / 3 } =& \omega^2 \\ e^{5 \pi i / 3 } =& - \omega \end{align*} $$ i.e., only when $u \in \left\{ \pm 1, \pm \omega, \pm \omega^2 \right\}$.

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