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I want to find coefficients $a$ and $b$ in $\mathbb{Z}[\sqrt{-19}]$ such that $a(18 + \sqrt{-19}) + b(18 - \sqrt{-19}) = 1$ (I know that these elements are coprime in $\mathbb{Z}[\sqrt{-19}]$).

At first I thought of the extended euclidean algorithm, but then I remembered that $\mathbb{Z}[\sqrt{-19}]$ is not an Euclidean domain, it's not even a unique factorization domain.

Setting $a = b = 1$ gives $a(18 + \sqrt{-19}) + b(18 - \sqrt{-19}) = 36$, but this does not help either because 36 is not a unit.

Is there a systematic way to go about this, or just guesswork?

Anakhand
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1 Answers1

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Let $z_1=18+\sqrt{-19}$, $z_2=18-\sqrt{-19}$. We have $$z_1+z_2=36$$ as well as $$ z_1z_2=343. $$ We have $\gcd(36,343)=1$, and a run of Euclid shows that $$1=162\cdot 36-17\cdot343.$$

Therefore $$ 1=162(z_1+z_2)-17z_1z_2=(162-17z_2)z_1+162z_2. $$

Jyrki Lahtonen
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    I first used the prime factors, but that lead to a more complicate solution. See the edit history if you are reaaaally interested :-) – Jyrki Lahtonen Sep 23 '23 at 17:41
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    I have no idea how generally this method works and even yields an answer in a non-maximal order of algebraic integers? – Jyrki Lahtonen Sep 23 '23 at 17:43
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    Probably some luck that "a run of Euclid" does succeed? That is, at least with a naive "norm", it could fail, despite the (correct) ring being a PID, but not norm-Euclidean? – paul garrett Sep 23 '23 at 17:45
  • I forgot that here the integral closure, while a PID, is not norm-Euclidean. Thanks for the reminder @paulgarrett. If you recall related more general results, please share. I guess I was lucky. – Jyrki Lahtonen Sep 23 '23 at 17:49
  • I capitalized on the fact that here the trace and the norm had no common factors. Of course it was essential that the two numbers happened to be each others conjugates. – Jyrki Lahtonen Sep 23 '23 at 17:56