# Finding all principal ideals of $\mathbb{Z}[\sqrt{-7}]$ containing a certain element.

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?

• Why do you assume these two elements have a gcd? Oct 9, 2020 at 1:02
• 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.
– ASP
Oct 9, 2020 at 1:34

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 $$\mathbb Z[\sqrt{-7}]$$.