Which prime numbers $p \in \mathbb{Z}$ are reducible in the unique factorization domain $\mathbb{Z}\left[\frac{1 + \sqrt{-3}}{2}\right]$ ?

Suppose $p$ is a prime integer and $p = \alpha \beta$ in $\mathbb{Z}\left[\frac{1 + \sqrt{-3}}{2}\right] = \mathbb{Z}[\omega]$. Then $f(p) = p^2 = f(\alpha)f(\beta)$, where $f: a + b \omega \mapsto a^2 + ab + b^2$ is the norm in $\mathbb{Z}[\omega]$.

There are only two possibilities:

  1. $f(\alpha) = 1, f(\beta) = p^2 $, so $\alpha $ is a unit
  2. $f(\alpha) = p, f(\beta) = p $

Set $\alpha = a + b \omega$, then $f(\alpha) = a^2 + b^2 + ab = p$. Therefore, if there exist integers $a$ and $b$ which solve this equation, then $p$ is reducible. And if this equation has no integer solutions, $p$ is prime in $\mathbb{Z}[\omega]$. But how can I express these solutions? I tried to use congruences $\text{mod} \ 4 $ , but it did not help much.


This is an example of where you want congruence modulo $3$ not $4$. There are two cases, either $p\equiv 1\mod 3$ or $p\equiv -1\mod 3$. Quadratic reciprocity says that

$$\left({p\over 3}\right)\left({3^*\over p}\right)=1$$

What we're looking for is $\left({3^*\over p}\right)$ since that tells us if $-3$ is a square mod $p$ and therefore if there is a solution.

But then this is just equal to $\left({p\over 3}\right)$ by multiplying both sides by $\left({p\over 3}\right)$. And we know this is

$$\left({p\over 3}\right) = \begin{cases} 1 & p\equiv 1\mod 3 \\ -1 & p\equiv -1\mod 3\end{cases}$$


Very minor detail, the function $f$ is more commonly notated $N$, for "norm."

More importantly, however, I think the norm function reveals itself more clearly if you regard it in this manner: given $a + b \sqrt{-3}$, for $\{a, b\} \in \mathbb{Z}$, then $N(a + b \sqrt{-3}) = a^2 + 3b^2$; or given $$\frac{a + b \sqrt{-3}}{2}$$ with $a$ and $b$ both odd integers, then $$N\left(\frac{a + b \sqrt{-3}}{2}\right) = \frac{a^2 + 3b^2}{4}.$$ Looking at $\omega$ has its uses, but at this point it tends to add a layer of confusion that blocks facility.

Then, for a prime $p$ to not be prime in this ring, it has to be such that $4p = a^2 + 3b^2$. So congruence modulo $4$ does not help us. Hence we need to look at congruence modulo $3$ instead. Since $4 \equiv 1 \pmod 3$, what we're looking for then is $p \equiv 1 \pmod 3$. That's because $a^2 + 3b^2 \equiv 2 \pmod 3$ is impossible.

Then, if you discover $4p = a^2 + 3b^2$ or $p = a^2 + 3b^2$, then you can just plug $a$ and $b$ into $a + b \sqrt{-3}$ or $$\frac{a + b \sqrt{-3}}{2}.$$ If you have to have it as $\alpha + \beta \omega$, from the form with halves do $\alpha = a + b$ and $\beta = b$ (the extra negative halves of $b \omega$ will be thus compensated -- if I did this correctly).


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