3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.