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I need to find all principal ideals in $\mathbb{Z}[\sqrt{-7}]$ that contain the ideal $(4,1+\sqrt{-7})$.

What I'm trying to do is find the $\gcd(4,1+\sqrt{-7})$, then this generator is one such ideal that I need, however I'm having trouble finding the gcd( . ). I'm doing the following,

$$\frac{4}{1+\sqrt{-7}}=\frac12-\frac12\sqrt{-7}.$$

Then I choose $1-\sqrt{-7}$ (I could've chosen [I think], $1$, $0$ or $-\sqrt{-7}$). Therefore,

$$4=(1-\sqrt{-7})(1+\sqrt{-7})+r.$$

Which implies that $r=-4$. One, the norm of $r$ didn't decrease, and no matter which choice of closest integer I chose the norm of $r$ never went below $N(1+\sqrt{-7})=8$. I'm not entirely sure what I'm doing wrong so any guidance would be appreciated.

EDIT: I've done some more work and found the following,

If we indeed have that $(4,1+\sqrt{-7})\subseteq(\alpha)$ for some $\alpha$. Then it must be that $N(\alpha)$ divides $8$. Meaning $N(a) \in \{1,2,4,8\}$. $1$ and $2$ are impossible. Then it is easy to check the only possible candidates for $\alpha$ is from the set $\{2,1+\sqrt{-7},1-\sqrt{-7}\}$. If $\alpha=1+\sqrt{-7}$ then there must exist some $\beta$ such that,

$$4=\beta(1+\sqrt{-7})$$

However $\beta$ has no solutions in $\mathbb{Z}[\sqrt{-7}]$ and we have a similar situation for for $1-\sqrt{-7}$. However now we are left with $\alpha=2$, so this implies that there exists some $\gamma$ such that,

$$1+\sqrt{-7}=2\gamma$$

However, this again has no solutions in $\mathbb{Z}[\sqrt{-7}]$. Does this mean that there doesn't exist any principal ideals that contain $(4,1+\sqrt{-7})$, or have I done something wrong?

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    $\begingroup$ Why do you assume these two elements have a gcd? $\endgroup$ Oct 9, 2020 at 1:02
  • $\begingroup$ I just thought that this was generally how I was going to find this principle ideal, as this was what I was taught in this particular unit. Otherwise I'm not particularly sure how to find these principal ideals. $\endgroup$
    – ASP
    Oct 9, 2020 at 1:34

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You've shown that, if $(4,1+\sqrt{-7})\subset (\alpha)$, then $N(\alpha)\in\{1,2,4,8\}$. One option is certainly $N(\alpha)=1$, which gives you the ideal $(1)$. You've correctly proven that none of $\{2,1\pm\sqrt{-7}\}$ divide both $4$ and $1+\sqrt{-7}$, so these do not work as $\alpha$. As a result, the only possible $\alpha$ is $1$ (or $-1$, I guess, but this gives the same ideal).

Generally speaking, the flaw in your original logic is that not integral domains are GCD domains -- so the $\gcd$ of two elements does not have to be defined. The condition of being a GCD domain is a bit weaker than that of PID or UFD (which $\mathbb Z[\sqrt{-7}]$ is not), but is stronger than being integrally closed (see here). In fact, $\mathbb Z[\sqrt{-7}]$ is not integrally closed: the roots $\frac{1\pm\sqrt{-7}}{2}$ of $x^2-x+2$ are algebraic elements of $\mathbb Q(\sqrt{-7})$, but are not in $\mathbb Z[\sqrt{-7}]$.

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