Efficiently computing GCDs in $\mathbb{Z}[(1+\sqrt{-19})/2]$

Question

The ring $\mathbb{Z}[(1+\sqrt{-19})/2]$ is a PID; hence any two elements have a GCD. How you would compute their GCD?

In a Euclidean domain, you would use the Euclidean algorithm. But $\mathbb{Z}[(1+\sqrt{-19})/2]$ is not Euclidean.

If you knew prime factorizations of the two elements, you could immediately compute the GCD. But surely factoring in $\mathbb{Z}[(1+\sqrt{-19})/2]$ is at least as hard as factoring in $\mathbb{Z}$. At least for now, we do not know how to do this efficiently.

So, is there an efficient algorithm to compute GCD?

(You may use the de facto interpretation of "efficient" as "polynomial time".)

Answer 1

Not a complete answer, just trying to draw attention to your question.

Humans such as yourself generally find factorization in $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ to be more difficult than factorization in $\mathbb{Z}$ because $\mathbb{Z}$ is familiar and $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is unfamiliar.

So if I was bedraggled by human failings, I would see about how to leverage my familiarity with one to my advantage in the other. The way is with norms.

If the norms of two numbers in $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ are coprime, then the two numbers must be coprime as well. Example: $$N\left( \frac{3 + \sqrt{-19}}{2} \right) = 7$$ and $$N\left( \frac{9 + \sqrt{-19}}{2} \right) = 25,$$ so since $\gcd(7, 25) = 1$, this must mean that $$\gcd\left( \frac{3 + \sqrt{-19}}{2}, \frac{9 + \sqrt{-19}}{2} \right) = 1$$ as well.

But if the norms do have a common divisor, then it's a little more difficult, for humans anyway. Example: $$\gcd\left( \frac{3 + \sqrt{-19}}{2}, 11 + \sqrt{-19} \right) = ???$$

Answer 2

The last time I wanted to do this, I used the $O(n^2)$ algorithm described in Agarwal and Frandsen. I think I recall that your specific case is easier because $-19 < 0$.