$p$ is prime and $p=a^2+b^2$ with $a,b\in \mathbb{Z}$. Show that $p=(a+ib)(a-ib)$ is a decomposition with prime elements in $\mathbb{Z}[i]$ and that $a+ib$ is associate with $a-ib$ if and only if $|a|=|b|=1$.

I've already proven that $(a+ib)$ and $(a-ib)$ are prime in the ring. I want to attempt the second statement.

It's clear to me that $(\pm1\pm i)$ is associate to $(\pm 1\pm i)$ so I'm just proving that if $a+ib$ is associate with $a-ib$ $\Rightarrow$ $|a|=|b|=1$

Definition: $a$ associate to $b$ if $\exists e\in R:$ $ae=b$ and $e$ is a unit in the ring $R$.

I know that there are $4$ units in total in $\mathbb{Z}[i]$, namely $1,-1,i,-i$.

\begin{align*} (a+ib)\cdot 1 = (a-ib) \Rightarrow b = 0\\ (a+ib)\cdot (-1) = (a-ib) \Rightarrow a = 0\\ (a+ib)\cdot i = (a-ib) \Rightarrow a = -b\\ (a+ib)\cdot (-i) = (a-ib) \Rightarrow a = b \end{align*} First two cases can't be true, since $p=(a+0)(a+0)=a^2$ and $p=(0+ib)(0-ib) = b^2$ wouldn't be prime.

For the two other cases we have $p=a^2+a^2=2a^2$ and that's only a prime if $|a|=|b|=1$.

Is that OK? Any feedback is welcome :)

  • 1
    $\begingroup$ Yes, your arguments are correct. $\endgroup$ – WhatsUp Jun 10 at 21:47

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