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I'm a bit confused at how to do this, I guess mostly because it is not a PID (or even a UFD). I'm interested in calculating the GCD of $(x + 2\sqrt{-5})$ and $(x - 2\sqrt{-5})$ for the same reason that this question was asked. However, I'm having trouble with this.

Firstly, they claim that they are relatively prime when $x$ is odd, but I don't see how this is always true. For example, let $x = 5$. If the ideals were relatively prime, there would exist elements $a + b \sqrt{-5}, c + d \sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that $$ (5 + 2\sqrt{-5})(a + b \sqrt{-5}) + (5 - 2\sqrt{-5})(c + d \sqrt{-5}) = 1. $$ Expanding this and looking at the equation of the integer part, we see $$ 5a - 10b + 5c + 10d = 1,$$ which obviously cannot be the case as the LHS is divisible by $5$ and the RHS is not. So the claim made in the other question is false. Now, we can see that $$ -2 (5 + 2\sqrt{-5}) + \sqrt{-5}(5 - 2\sqrt{-5}) = \sqrt{-5},$$ and $(\sqrt{-5})$ is a prime ideal (and is thus maximal as $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain).

Fnark Man
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  • About gcd in this ring, see this post, for example. – Dietrich Burde Apr 13 '25 at 18:21
  • @DietrichBurde Hmm, that's definitely very helpful for understanding, but I'm not sure if I can calculate the GCDs that I have here in general. I feel like doing it for a general $x$ as I want is really hard, but seems to be necessary. – Fnark Man Apr 13 '25 at 20:20

1 Answers1

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You are right that calculating GCDs outside of a PID or UFD is more subtle. In fact, it's not clear that they even exist, like in the post linked in the first comment shows. There is also a typo in this other post: $(5 + 2\sqrt{-5}), (5 - 2\sqrt{-5})$ have $\sqrt{-5}$ as a common divisor and are thus not relatively prime. The claim is true for $5 \nmid x$, which could be assumed in the original problem since $125x^3 - 20$ is divisible by $5$ by not $25$ and is thus never a square.

The key idea is that $R = \mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain, which means (by definition) that even though $R$ may fail to be a UFD, every ideal in $R$ still factors uniquely as a product of prime ideals.

Thus every pair of principal ideals $(a), (b)$ in $R$ has an ideal GCD $(a,b) \subseteq R$, and then $(a), (b)$ have a GCD if and only if $(a,b)$ is principal, in which case their GCD is the principal generator of $(a,b)$.

For instance if $x$ is odd and not equal to $5$, we have that the ideal GCD of the ideals $(x - 2\sqrt{-5}),(x + 2\sqrt{-5})$ is equal to $$(x - 2\sqrt{-5}, x + 2\sqrt{-5}) \subseteq (x^2 - 20, 4\sqrt{-5}) \subseteq (x^2-20, 20)$$ The GCD of $(x^2- 20, 20)$ in $\mathbb{Z}$ is $1$ since $x$ is not divisible by $2$ or $5$, so their GCD in $R$ is also $1$.

KCd
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emilg
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