$\mathbb{Z}[\sqrt{-7}]$ is not an Euclidian Domain I want to prove that $\mathbb{Z}[\sqrt{-7}]$ is not an Euclidian Domain.
First, I know that Euclidian  Domain $\implies$ principal, i.e., every ideal is principal. So, I want to construct a ideal in $\mathbb{Z}[\sqrt{-7}]$ that is not generated by a unique element $a+b\sqrt{-7}\in\mathbb{Z}[\sqrt{-7}]$, $a,b\in\mathbb{Z}$.
I saw here the ideal $I=(2,1+\sqrt{-7})$ is not principal, but the answers uses some Galois Theory, and I haven't learned it yet.
My questions are:


*

*Where did the construction of this ideal come from? What is the motivation?

*How to prove that $I$ is not principal?

 A: $\mathbb{Z}[\sqrt{-7}]$ isn't a UFD so in particular it couldn't be a Euclidean domain.
So let's study the element $8$. $$(1+\sqrt{-7})×(1-\sqrt{-7})=8=2×2×2.$$We have to show $2,1+\sqrt{-7},1-\sqrt{-7}$ are irreducibles. 
Let study $1+\sqrt{-7}=(a+\sqrt{-7}b)(c+\sqrt{-7}d)$ $a,b,c,d\in \mathbb{Z}$. Taking seminorms $(a^2+7b^2)(c^2+7d^2)=8\implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $d\not=0\implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2\geq 28$. Also $c\not=0$ , as $7\not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2\geq 11$. Therefore $a=1,-1$ i.e. $a+\sqrt{-7}b$ is unit. 
Now let $2=(a+b\sqrt{-7})(c+d\sqrt{-7})$ such that $a,b,c,d \in \mathbb{Z}$, then pass at the semi-norms :$$(a^2+7b^2)(c^2+7d^2)=4\implies b=d=0$$ $e,f\in \mathbb{Z}$ and wlog (because in $\mathbb{Z}$, $2$ is irreducible)  $$f\not =0\implies e^2+7f^2\geq 7$$ Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $\Bbb Z$ is a UFD i.e. either $a+b\sqrt{-7}$ or $c+d\sqrt{-7}$ unit. 
In the same way $1-\sqrt{-7}$ is irreducible.
A: The question becomes easier if we make it about inclusions of ideals of $\Bbb{Z}$, i.e. about divisibility of integers. To do this, all you need to know is that the norm function
$$\mathcal{N}:\ \Bbb{Z}[\sqrt{-7}]\ \longrightarrow\ \Bbb{Z}:\ a+b\sqrt{-7}\ \longmapsto\ (a+b\sqrt{-7})(a-b\sqrt{-7})$$
is multiplicative. That is to say, for $x,y\in\Bbb{Z}[\sqrt{-7}]$ you have $\mathcal{N}(xy)=\mathcal{N}(x)\mathcal{N}(y)$.
Now suppose the ideal $(2,1+\sqrt{-7})\subset\Bbb{Z}[\sqrt{-7}]$ is principal, say it is generated by $x\in\Bbb{Z}[\sqrt{-7}]$. Then
$$2=xy\qquad\text{ and }\qquad 1+\sqrt{-7}=xz,$$
for some $y,z\in\Bbb{Z}[\sqrt{-7}]$. It follows that
$$\mathcal{N}(x)\mathcal{N}(y)=\mathcal{N}(xy)=\mathcal{N}(2)=4,$$
which shows that $\mathcal{N}(x)$ divides $4$. This leaves only a few cases to check:


*

*If $\mathcal{N}(x)=\pm1$ then $x$ is a unit and so 
$$(2,1+\sqrt{-7})=(x)=\Bbb{Z}[\sqrt{-7}].$$
Show that $(2,1+\sqrt{-7})$ is a proper ideal to arrive at a contradiction.

*If $\mathcal{N}(x)=\pm2$ then writing $x=a+b\sqrt{-7}$ we find that
$$\pm2=\mathcal{N}(x)=\mathcal{N}(a+b\sqrt{-7})=a^2+7b^2.$$
Reducing mod $8$ yields a contradiction.

*If $\mathcal{N}(x)=\pm4$ then $\mathcal{N}(y)=\pm1$ and so $y$ is a unit, so
$$(2,1+\sqrt{-7})=(x)=(2),$$
but it is clear that $1+\sqrt{-7}\notin(2)$, a contradiction.


This shows that the ideal is not principal.
