Eisenstein Prime Number Theorem Proof
Theorem
An Eisenstein prime is an irreducible element of the Eisenstein ring. An Eisenstein integer $\pi \in \mathbb{Z}[ \omega ]$ is an Eisenstein prime if it satisfies one of the following conditions:
- (i): $\pi = 1 + \omega 2$
- (ii): For some prime $p \in \mathbb{Z}$, $p \equiv 2 \pmod{3}$ of $\pi = p$
- (iii): For some prime $p \in \mathbb{Z}$, when $p \equiv 1 \pmod{3}$, $p = u^2 - uv+ v^2$ is satisfied by $\pi = u + \omega v$
- (iv): $ \pm \omega^{k} \pi$ obtained by multiplying the unit $\pm 1 , \pm \omega , \pm \omega^2$ of $\mathbb{Z} [\omega ]$ to $\pi$ corresponding to (i)~(iii)
- (v): $\overline{\pi}$ obtained by taking the conjugate of $\pi$ corresponding to (i)~(iii)
Description
When talking about Eisenstein integers, $\pi$ usually refers to Eisenstein primes, not pi, to avoid confusion with the commonly used term Natural Prime $p$ for regular integer primes. This extension of primes has made studies on Eisenstein integers analogous to number theory, similar to the case with Gaussian primes.
(i)
$(1 + \omega 2)$ acts as a substitute for $3 = - (1 + \omega 2)^2$, essential for the factorization of $3 \in \mathbb{Z} [ \omega ]$. Although prime $2 \in \mathbb{Z}$ is of type (ii) and thus cannot be considered the smallest prime, the same was true in the context of $\mathbb{Z}$.
(ii)
For example, $5$ cannot be factorized using Eisenstein integers. It’s pointless to check if this is truly impossible, so it’s recommended to just look at the proof.
(iii), (iv), (v)
For example, since $7$ is $7 = (3 + \omega)(2 - \omega )$, it can be factorized. Here, $\mathbb{Z}[ \omega ]$ is a UFD, hence it has a unique factorization. Meanwhile, $(3 + \omega)$ forms a conjugate with $2 - \omega = 3 - (1 + \omega) = 3 + \overline{\omega}$ making it an Eisenstein prime.
Proof
Strategy: $(Z[ \omega ] , N)$ is the norm of the Eisenstein ring defined as $N(x+\omega y) = x^2 - xy + y^2$ (where $x, y \in \mathbb{Z}$). Though the proof of the Eisenstein prime theorem merely brings algebraic properties together with various results from elementary number theory, understanding these properties and results is the challenging part.
Part 0. If $| N( \pi ) | = p$ is a prime, $\pi$ is an Eisenstein prime.
Multiplicative norm property: Let’s say $ p \in \mathbb{Z}$ is a prime.
- [1]: If the multiplicative norm $N$ is defined in $D$, then $N(1) = 1$ and for all units $u \in D$, $| N ( u ) | = 1$
- [2]: If all $\alpha \in D$ satisfying $| N ( \alpha )| =1$ are units in $D$, then $\pi \in D$ satisfying $| N ( \pi ) | = p$ is an irreducible element in $D$.
Properties of the Eisenstein ring
- [3]: The only units in $\mathbb{Z}[\omega]$ are $\pm 1, \pm \omega , \pm \omega^2$.
According to [3], the only units in $\mathbb{Z}[ \omega ]$ are $\pm 1, \pm \omega , \pm \omega^2$, and by [1], $N(\pm1) = N(\pm \omega) = N( \pm \omega^2 ) = 1$ holds. Since all $\alpha$ satisfying $| N ( \alpha )| =1$ were units in $\mathbb{Z}[ \omega ]$, by [2], $\pi$ that satisfies $| N( \pi ) | = p$ becomes an irreducible element in $\mathbb{Z}[ \omega ]$. In other words, if $| N( \pi ) | = p$ is a prime, $\pi$ is an Eisenstein prime.
Part (i). $\pi = 1 + \omega 2$
If $\pi = 1 + \omega 2$, then $N(\pi) = 1^2 - 1 \cdot 2 + 2^2 = 3$ is a prime, making $\pi = 1 + \omega 2$ an Eisenstein prime according to Part 0.
Part (ii). $\pi \equiv 2 \pmod{3}$
Suppose $\pi = p$ satisfies $p \equiv 2 \pmod{3}$ for a prime of $\mathbb{Z}$ but isn’t a Gaussian prime in $\mathbb{Z}[ \omega ]$, hence it has a factorization like $\pi = ( a + \omega b )( c + \omega d )$. By the multiplicative property of $N$, $$ \begin{align*} p^2 =& \pi^2 - \pi \cdot 0 + ^2 \\ =& N ( \pi + \omega 0) \\ =& N ( a + \omega b ) N ( c + \omega d ) \\ =& (a^2 - ab+ b^2) (c^2 - cd + d^2) \end{align*} $$ We get $p^2 = (a^2 - ab + b^2) (c^2 -cd + d^2)$, where since $p \in \mathbb{Z}$ is a prime, there must exist a solution satisfying $\begin{cases} a^2 - ab + b^2 = p \\ c^2 - cd + d^2 = p \end{cases}$.
Necessary and sufficient condition for a prime to leave a remainder of 1 when divided by 3: If $p \ne 3$ is a prime, and $p \equiv 1 \pmod{3}$ $\iff$, for some $a,b \in \mathbb{Z}$, $p = a^2 - ab + b^2$
However, since $p \equiv 2 \pmod{3}$, there does not exist a solution satisfying $\begin{cases} a^2 - ab + b^2 = p \\ c^2 - cd + d^2 = p \end{cases}$ according to the necessary and sufficient condition for a prime to leave a remainder of 1 when divided by 3, leading to a contradiction, thus $\pi \equiv 2 \pmod{3}$ is a Gaussian prime.
Part (iii). $\pi = u + \omega v$
For a prime $p \in \mathbb{Z}$, since $p \equiv 1 \pmod{3}$, according to the necessary and sufficient condition for a prime to leave a remainder of 1 when divided by 3, $$ N (\pi) = N ( u+ \omega v) = u^2 -uv + v^2 = p $$ $\pi = u + \omega v$ satisfying this condition is a Gaussian prime according to Part 0.
Part (iv). $ \pm \omega^{k} \pi$
Since Part 0 declared $\pi$ an Eisenstein prime if $| N( \pi ) | = p$ is a prime, then for $k \in \mathbb{Z}$, $$ \begin{align*} N( \pm \omega^{k} \pi ) =& N( \pm \omega^{k} ) N (\pi ) \\ =& 1 \cdot N (\pi ) \\ =& p \end{align*} $$ $\pm \omega^{k} \pi $ satisfying this also are Eisenstein primes.
Part (v). $\overline{\pi}$
Taking the conjugate of an Eisenstein integer, $$ \begin{align*} \overline{x + \omega y} =& x + \overline{\omega} y \\ =& x - (1 + \omega) y \\ =& (x-y) - \omega y \end{align*} $$ and since Part 0 declared $\pi = x + \omega y$ an Eisenstein prime if $| N( \pi ) | = x^2 - xy + y^2 = p$ is a prime, $$ \begin{align*} N( \overline{\pi} ) =& N \left( (x-y) - \omega y \right) \\ =& (x-y)^2 - (x-y) \cdot (-y)+ (-y)^2 \\ =& x^2 - 2 xy + y^2 + xy - y^2 + y^2 \\ =& x^2 - xy + y^2 \end{align*} $$ $\overline{\pi}$ satisfying this also are Gaussian primes.
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