Definition of gcd in Euclidean domain dependent on valuation? Let $D$ be Euclidean. In $\mathbb Z[i]$ our lecturer defined  $\gcd (a,b)$ as the common divisor with greatest norm. I have a slight problem with this, since we actually fix the norm $N(a+bi)=a^2+b^2$.
Does this definition indeed depend on the norm or is it actually equivalent to the general definition via the universal property $$c\mid \gcd(a,b)\iff c\mid a,b$$
 A: First note that there are several $gcd$s of $a$ and $b$, so it should be "a common divisor with maximal norm".
In general euclidian domains
In general it is true that you can define a $gcd$ as a common divisor with maximal euclidian "norm", if and only if your norm satisfies the condition :
$(*)$ if $a|b$ then $N(a)\leqslant N(b)$, with equality if and only if $a$ and $b$ are associated. 
I do not know if you can always choose such a norm ; it is true (see https://en.wikipedia.org/wiki/Euclidean_domain ) that you can always choose one such that $a|b \implies N(a)\leqslant N(b)$ but I don't know about this stronger condition.
If $(*)$ is not satified, then take any counterexample : either $a$ and $b$ are associated but $N(a)>N(b)$, or $a$ strictly divides $b$ but $N(a)\geqslant N(b)$. In both cases $b$ is certainly a $gcd$ of $b$ and $b$. But in the first case, $b$ does not have maximal norm among its divisors. In the second case, if $b$ has maximal norm among its divisor, then so does $a$ but it is not a $gcd$, and if not then there is a $gcd$ that does not have maximal norm. In all cases, the definition of $gcd$s as common divisors with maximal norm is broken.
Conversely, if $(*)$ is satisfied, then if $a,b\in R$, then the common divisors of $a$ and $b$ are precisely the $x\in R$ such that $(a)+(b)\subset (x)$. But there is a $d$ such that $(a)+(b)=(d)$ since $R$ is principal, so the common divisors are precisely the $x$ such that $(d)\subset (x)$, ie $x|d$. For now I have not used the norm except to show that $R$ is principal (which is independent of the norm). So now $(*)$ gives that among divisors $x$ of $d$, those with maximal norm are precisely those such that $(x)=(d)$, so the $gcd$s of $a$ and $b$.
Case of a number field
But in your case it is way easier, because your norm is not just any made-up euclidian function, it is a real, honest norm, as in "the norm of a number field". So it satisfies the very convenient property that if $a|b$ then $N(a)|N(b)$, and $N(a)=1$ if and only if  $a$ is a unit, which obviously is way stronger than $(*)$, and makes the proof simpler.
