For any $p,q\in\mathbb{Z}[i]$, $\mathrm{N}(\gcd(p, q))$ must divide $\gcd(\mathrm{N}(p), \mathrm{N}(q))$

Question

I'm studying for my final exam (abstract algebra) and am looking at an example where our professor was trying to compute the GCD of two elements of $\mathbb{Z}[i]$. Rather than directly applying the Euclidean algorithm, he used the fact that (denoting the respective elements of $\mathbb{Z}[i]$ as $p$ and $q$) the norm of the gcd of $p$ and $q$ must divide the gcd of the norms of $p$ and $q$. Formally,

$$ \mathrm{N}(\gcd(p, q)) \; \text{must divide} \; \gcd(\mathrm{N}(p), \mathrm{N}(q)). $$

I'm looking through our textbook (Dummit & Foote) and can't seem to find this anywhere. Could anyone give me an explanation for why this must be true (proof or just intuitive reasoning)?

Answer 1

Hint $\ \ d\mid p,q\,\Rightarrow\, \bar d\mid \bar p,\bar q\Rightarrow\, d\bar d\mid p\bar p,q\bar q\overset{\rm\color{#c00}U}\Rightarrow\, d\bar d\mid(p\bar p,q\bar q)\,$ by ${\rm\color{#c00}U}$ = univ. gcd property

Remark $\ $ The proof works in any UFD/GCD domain, for $\, x\mapsto \bar x$ generalized from conjugation to any multiplicative map, which necessarily preserves divisibility, enabling the first arrow above.

Answer 2

By definition, we have $\gcd(p,q)\mid p$ and $\gcd(p,q)\mid q$.

For any $r,s\in\mathbb{Z}[i]$, if $r\mid s$ then by definition $s=kr$ for some $k\in\mathbb{Z}[i]$, hence $\mathrm{N}(s)=\mathrm{N}(k)\mathrm{N}(r)$, so we must have $\mathrm{N}(r)\mid \mathrm{N}(s)$.

Therefore, $\mathrm{N}(\gcd(p,q))\mid \mathrm{N}(p)$ and $\mathrm{N}(\gcd(p,q))\mid \mathrm{N}(q)$.

But by definition, if $a,b,c\in\mathbb{Z}$ and $a\mid b$ and $a\mid c$, then $a\mid \gcd(b,c)$.

Therefore, $\mathrm{N}(\gcd(p,q))\mid \gcd(\mathrm{N}(p),\mathrm{N}(q))$.

Note that there need not be equality here. For example, with $p=1+2i$ and $q=1-2i$, $$\mathrm{N}(\gcd(p,q))=\mathrm{N}(1)=1\qquad \gcd(\mathrm{N}(p),\mathrm{N}(q))=\gcd(5,5)=5$$