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We were given a hint to set $p$ equal to the factorization of $p$ into two elements then to take the norm of both sides. Here is what I've done so far:

Since $p$ is not prime in $Z[i]$, we can rewrite $p = (a + bi)(c + di)$ for some $a, b, c, d \in Z$. So we can take the norm of both sides to get

$\sqrt{p\overline{p}} = \sqrt{(a + bi)(c + di)\overline{(a + bi)(c + di)}} = \sqrt{(ac + adi + bci - bd)\overline{(ac + adi + bci - bd)}} = \sqrt{((ac - bd) + (ad + bc)i)\overline{((ac - bd) + (ad + bc)i)}} = \sqrt{((ac - bd) + (ad + bc)i)((ac - bd) - (ad + bc)i)} = \sqrt{(ac-bd)^2 + (ad + bc)^2}$.

This is where I'm stuck. I've tried reversing the norm at this last step but it basically just puts me back right at the beginning. I've recognized that $\sqrt{p\overline{p}} = p$ but I'm not sure if I can do anything with this.

Juniven Acapulco
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UBears
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2 Answers2

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So, $p$ is composite in $\Bbb Z[i]$ so that $p=\alpha\beta$ for some $\alpha,\beta\in\Bbb Z[i]$ where $1<N(\alpha)<N(p)$ and $1<N(\beta)<N(p)$. Hence, $$p^2=N(\alpha)N(\beta).$$ This implies that $$N(\alpha)\mid p^2.$$ Because $p$ is prime, either $N(\alpha)=1$ or $p$ or $p^2$. We cannot have $N(\alpha)=1$ and $N(\alpha)=p^2$. Hence, $N(\alpha)=p$. Write $\alpha=a+bi$ and we are done.

Juniven Acapulco
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  • Thank you for your response, I think I understand most of it. I'm just a little confused how you get to $p^2 = N(\alpha)N(\beta)$. I wonder if I may have been computing the norm incorrectly. I have been doing $N(z) = \sqrt{z\overline{z}}$. For a real value like $p = 13 + 0i$ for example, I would get $N(p) = N(13) = \sqrt{(13 + 0i)(\overline{13 + 0i})} = \sqrt{13^2} = 13 = p$. I'm unsure how this would lead to $p^2 = N(\alpha)N(\beta)$? – UBears May 10 '17 at 00:31
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    @UBears Just take norm to both sides of $p=\alpha\beta$. For the definition of the norm, please read this – Juniven Acapulco May 10 '17 at 00:36
  • I understand how taking the norm of $\alpha \beta$ gives $N(\alpha \beta) = N(\alpha)N(\beta)$ but I'm confused how taking the norm of $p$ gives $N(p) = p^2$. Wouldn't it give $N(p) = p$ like I calculated in my previous comment? – UBears May 10 '17 at 00:42
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    @UBears Note that the norm of $z\in\Bbb C$ is given by $N(z)=|z|^2$ which means that if $z=a+bi$ then $N(z)=a^2+b^2$. In particular, $$N(p)=N(p+0i)=p^2+0^2=p^2.$$ – Juniven Acapulco May 10 '17 at 00:47
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Suppose that a natural prime number $p$ is not a Gaussian prime. As $\mathbb{Z}[\text{i}]$ is a unique factorization domain, $p$ is divisible by an irreducible element $\pi\in\mathbb{Z}[\text{i}]$. Suppose that $p=\pi\varpi$ for some $\varpi\in\mathbb{Z}[\text{i}]$. Show that $N(\pi)=N(\varpi)=p$, where $N(z):=|z|^2$ for every $z\in\mathbb{C}$. This immediately implies that $\varpi=\bar\pi$. If $\pi=a+b\text{i}$ for some $a,b\in\mathbb{Z}$, then $p=\pi\bar\pi=a^2+b^2$.

Batominovski
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