$\mathbb{Z}[\sqrt{-7}]$ is not an Euclidian Domain

Question

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?

Note that norms don't require any knowledge of Galois theory, so the arguments in your linked post should be accessible to you. The whole idea is to move certain questions of divisibility in your ring to divisibility of integers (the norms) — Brian Moehring, Sep 05 '19 at 12:23
@BrianMoehring can you explain the first answer for me, please? — Mateus Rocha, Sep 05 '19 at 12:39
There is no Galois theory used in any of the linked answers (only a joke in a comment about user named Galois). If you have questions about a particular answer then you should post them in comments on the answer. — Bill Dubuque, Sep 05 '19 at 13:27

score 3 · Answer 1 · answered Sep 05 '19 at 12:44

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

Servaes · Accepted Answer · 2019-09-05T13:06:22.797

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.

I only have one question about 2.
"Reducing mod 8 yields a contradiction". Could you explain me? — Mateus Rocha, Sep 05 '19 at 14:59

$\mathbb{Z}[\sqrt{-7}]$ is not an Euclidian Domain

2 Answers2