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Examples:

$$\gcd(-9 - 3i, -2 + i)$$

$$\gcd(9 + 3i, -2 - i)$$

$$\gcd(-9 - 3i, -2 - i)$$

$$\gcd(-3 - 9i, -2 + i)$$

$$\gcd(-3 + 9i, -1 + 2i)$$

I've put these through Wolfram Alpha and, for some, gotten exactly the result I expected, but for some I have been surprised by the results, but I'm well aware that WA can in some cases get confused and arrive at the wrong result.

For justification of the answers, however nonsensical they may seem, I'm more interested in something that follows logically and by commonsense from analogy to $\mathbb{Z}$ and the Euclidean algorithm, rather than something that involves some exotic mathematical object presented as part of an appendage measuring contest.

Bonus question: how neatly does this carry over to other quadratic imaginary rings that are Euclidean? What about reals?

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    $\begingroup$ I think the ambiguity is unavoidable. $\endgroup$ – paul garrett Aug 11 '15 at 21:28
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    $\begingroup$ Counter-intuitive as it may seem, $$\frac{1 + 2i}{-1 + 2i} = \frac{3}{5} - \frac{4i}{5} \not\in \mathbb{Z}[i].$$ If you're very dyslexic, the Gaussian integers can be very confusing. $\endgroup$ – Robert Soupe Aug 12 '15 at 2:36
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Is the $\gcd$ of $12$ and $8$ equal to $4$ or to $-4$?

  • It should be a divisor of both $8$ and $12$.
  • It should also not divide any other number that divides both $8$ and $12$.

$4$ does divide another number that divides both $8$ and $12$, namely $-4$, and $-4$ does divide another number that divides both $8$ and $12$, namely $4$.

The two desiderata above need some revision once we work outside the set of positive integers. What is done is this: $4$ is considered equivalent to $-4$ because there is a "unit" by which $4$ can be multiplied to get $-4$, and that "unit" is $-1$. A "unit" is a divisor of $1$. Within the Gaussian integers, there are four units: $\pm1,\pm i$. All four are divisors of $1$ within the Gaussian integers. There is no Gaussian integer by which $3$ can be multiplied to get $1$, so $3$ is not a "unit".

So $1+2i$, $-2+i$, $-1-2i$, $2-i$ are all equivalent because any of them can be multiplied by one of the four "units" to get one of the others.

In that sense, it doesn't matter which is reported to be the $\gcd$.

However, if one were to adopt a convention that the answer should always be in the first quadrant and not a pure imaginary, then one has a "canonical form", and canonical forms provide a way to tell whether two things are equal: they're equal if they're identical once reduced to the canonical form. Thus for example one puts simple radical expressions into "simplest radical form" and then one can tell whether two of them are equal.

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  • $\begingroup$ Yes the concept of canonical form is under-taught. $\endgroup$ – user21820 Aug 20 '15 at 16:04
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Every gaussian prime $z$ has four associates

$$z, iz, -z, -iz$$

There will always be exactly one of those associates in the first quadrant or on the positive real axis.

If you want some sort of uniqueness to prime factorization, then you could do it this way.

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