Prove $(5-i,13)$ is a principal ideal in $\mathbb{Z}[i]$

Question

I'm doing the same exercise as the one asked about in this post. The only part I was not able to solve (the rest of the exercise is not relevant for this part) is to prove that $(5-i,13)$ is a principal ideal and can thus be written as $(a)$ for an $a\in\mathbb{Z}[i]$.

I am aware of the Euclidian Algorithm and already noticed that $13=(3-2i)(3+2i)$, and that $(5-i)/(3+2i)=(1-i)$. I don't know hoe to continue now and find a single generator for $(5-i,13)$. Can anyone provide a hint?

By the way, I cannot use the fact that any Euclidian ring is a principal ideal domain.

Tsemo Aristide · Accepted Answer · 2018-09-25T08:32:06.907

3

$3+2i=13-2(5-i)$, implies that $3+2i\in (13,5-i)$. This implies that $(3+2i)\subset (13,5-i)$. Since $13=(3-2i)(3+2i)$ and $5-i=(1-i)(3+2i)$, you deduce that $(13,5-i)\subset (3+2i)$.

edited Sep 25 '18 at 08:32

answered Sep 25 '18 at 08:25

Tsemo Aristide

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Wow, I don't get why I didn't see this... Thank you! – Václav Mordvinov Sep 25 '18 at 08:30
One comment: $5 - i = (1 - i)(3 + 2i),$ not $13(1 - i)(3 + 2i).$ – Stahl Sep 25 '18 at 08:31

Prove $(5-i,13)$ is a principal ideal in $\mathbb{Z}[i]$

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