The prime elements of the ring $\mathbb{Z} \! \left[ \dfrac{i \sqrt{3} - 1}{2} \right]$.

I have to study the prime elements of the ring $\mathbb{Z} \! \left[ \dfrac{i \sqrt{3} - 1}{2} \right]$. For the moment, I cannot find the general form of such elements. Can you help me?

Thanks! :)

• What is $j^2$ ? – Hurkyl Jun 17 '14 at 5:32
• For information and references, please see this Wikipedia article on Eisenstein primes. – André Nicolas Jun 17 '14 at 5:35
• A little hint.... There are two types of primes in $Z$: 4k+1 and 4k+3 Check and see in which structure your j can come in handy... – Nir Agami Jun 17 '14 at 5:36

There are two collections of primes in this ring:

$$p, pj, \mbox{ and } pj^2$$ where $p \in \Bbb{Z^+}$ is a prime of the form $3k+2$, and $$a + bj$$ where $a, b \in \Bbb{Z^+}$ and $a^2 -ab + b^2$ is a prime of the form $3k+2$.

All such numbers are prime and those are the only primes. I think Euler showed this first.

• +1, but I wanted to add that 1) $i\sqrt3$ is also a prime, 2) $-1$ is also a unit giving more associates. – Jyrki Lahtonen Jun 17 '14 at 5:58
• Don't you mean $a^2-ab+b^2$ of the form $3k+1$ or equal to 3? – user10676 Jun 17 '14 at 6:17
• For user10676: You are right, I should have said $a^2 - ab + b^2$ must be a prime not of the form $3k + 2$. Doh! – Mark Fischler Jun 17 '14 at 22:29
• o Jyrki -- you are half right. I should have said $\pm p, \pm pj, \pm pj^2$. But $i\sqrt{3}$, while prime, falls into the second class, with $a = 1, b=2$ and $a^2 -ab + b^2 = 3$ which is prime. Again, my mis-wording about the prime bing $3k+2$ may have led to this confusin. – Mark Fischler Jun 17 '14 at 22:34

These are the Eisenstein primes, and it is more common to use the notation $\mathbb{Z}[\omega]$ where $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2} = \frac{-1 + \sqrt{-3}}{2}.$$ (This is a very special number for reasons that may seem like a sidetrack from your question).

According to MathWorld, http://mathworld.wolfram.com/EisensteinPrime.html there are three classes of Eisenstein primes:

• $1 - \omega$ all by itself, apparently
• A positive prime $p \in \mathbb{Z}$ which also satisfies $p \equiv 2 \mod 3$ is also prime in $\mathbb{Z}[\omega]$. (If instead $p \equiv 1 \mod 3$, it can be divided by Eisenstein integers of smaller norms; for example, try factoring $7$ or $13$).
• A number of the form $a + b \omega$ or $a + b \omega^2$ such that $a^2 - ab + b^2 = p \equiv 1 \mod 3$ is a positive prime $p \in \mathbb{Z}$; therefore that prime number is actually composite in $\mathbb{Z}[\omega]$ and one of its factors is $a + b \omega$ or $a + b \omega^2$ (which is what I was getting at with my previous parenthetical).

If you have Mathematica, I would strongly recommend going to http://mathworld.wolfram.com/EisensteinInteger.html and downloading that Mathematica notebook to play around with these numbers a little bit. If you don't have Mathematica, try factoring some small positive integers in $\mathbb{Z}[\omega]$, e.g., 19, 23, 27, 29, 31, etc.