📂Number Theory

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.

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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.

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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.

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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.

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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.

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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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