# Prove that every ideal of a Euclidean domain is principal

Prove that every ideal of a Euclidean domain is principal.

I'm new to algebra, so the solution below is probably really awkward and unconcise.

So I was given the hint that I should let $$I$$ be an ideal and consider when $$I= \{0\}$$ and when $$I\neq \{0\}.$$ Obviously, $$I=\{0\}=\langle 0\rangle$$ is principal. As for $$I\neq \{0\},$$ I think I need to think of the smallest/minimal nonzero element, say $$a$$, in $$I$$ and show that $$I=\langle a\rangle.$$ As well, I know that an integral domain $$D$$ is a Euclidean domain iff for all $$a,b\in D,$$ $$a=bq+r,$$ where $$b\neq 0,$$ and $$r=0$$ or $$N(r) where $$N(r)$$ denotes the norm of $$r.$$ That is, from my understanding, $$N(r)$$ is the function $$N : R \to \mathbb{N}\cup \{0\}$$ such that $$N(0)=0$$ (am I forgetting anything here?) I just need to show that $$\forall x\in I\Leftrightarrow x\in \langle a\rangle.$$ Let $$x,y\in I$$ and $$r=x-ya.$$ If $$r=0,$$ we are done. If $$r\neq 0,$$ then by the well-ordering principle, there is a minimum value of $$r,$$ say $$r_1.$$ But $$a$$ is the minimum value of $$I,$$ so $$r_1 which means that $$r_1$$ must be zero and thus $$I=\langle a\rangle.$$ I think I'm missing something here.

• It makes sense for you to compare your proof with the standard proof first, see for example here, and then improve your own proof. – Dietrich Burde Oct 30 '19 at 19:10
• @DietrichBurde thanks I really need to improve my abstract algebra skills. I'm new to the field. – user718615 Oct 30 '19 at 19:11
• Did you show that $\Bbb{Z}$ is a PID using that for $|a| \ge| b|> 0$ there exists $c$ such that $(a,b) = (b,a-cb)$ with $|b-ca|< |b|$ ? The point of Euclidean domain is to do the same replacing $|.|$ by $N(.)$. – reuns Oct 30 '19 at 19:14
• @reuns why isn't it $|a-bc| < |b|?$ I got that bc $a=bc+r$ and $r<|b|$ – user718615 Oct 30 '19 at 20:05
• sure I meant $|a-bc|<|b|$ – reuns Oct 30 '19 at 20:10

The argument is generally okay (it proceeds along the usual lines), but with some small errors.

You should not write $$\langle a\rangle = \{r_1a, r_2a,\ldots, r_na\mid r_i\in R\}$$, because that implies that $$R$$ is finite. Rather, you should write $$\langle a \rangle = \{ra\mid r\in R\}$$.

(This holds in this case because we are assuming that $$R$$ is commutative and has a unit; if either of those things does not hold, the description of the principal ideal generated by $$a$$ is slightly more complicated).

The way you wrote the Euclidean property is also not quite right. Rather, the condition that $$b\neq 0$$ precedes the equation: if $$b\neq 0$$, then for all $$a$$ there exists $$q,r\in D$$ such that $$a=qb+r$$ and $$r=0$$ or $$N(r)\lt N(b)$$. (Of course, you also need the existence of the function $$N$$...)

In general you do not require $$N(0)=0$$; for example, the degree function on $$\mathbb{R}[x]$$ makes the ring $$\mathbb{R}[x]$$ into a polynomial ring, but we do not have that the $$0$$ polynomial has degree $$0$$; the degree is usually either undefined or called "$$-\infty$$". The Euclidean function is required to be defined only on $$D-\{0\}$$. The defining property of $$N$$ is that if $$ab\neq 0$$ and $$b\neq 0$$, then $$N(ab)\geq N(b)$$, and that the division algorithm holds.

You can't compare $$r_1$$ with $$a$$: $$a$$ is in $$R$$, whereas $$r_1$$ is an integer. Rather, you should have that you cannot have $$r_1\lt N(a)$$. That is what implies that $$r_1=0$$. But in any case, your derivation of $$r_1$$ doesn't work.

You want to show that $$I\subseteq \langle a\rangle$$. So let $$x\in I$$. Then, because $$a\neq 0$$, there exists $$q,r\in R$$ such that $$x=aq+r$$ and either $$r=0$$ or $$N(r)\lt N(a)$$. Since $$a$$ is chosen from among all elements of $$I$$ so that $$N(a)$$ is minimal, we cannot have $$N(r)\lt N(a)$$. So that means that $$r=0$$, and hence $$x=aq\in\langle a\rangle$$.

• thanks for telling me the important things that I got wrong! – user718615 Oct 30 '19 at 20:06
• my biggest problem I find is that I can recall the definitions okay, but I waste time figuring out how to use the definitions to solve problems. Anyways @ArturoMagidin, would you mind explaining a few things for me? For one, why didn't you prove that $\langle a\rangle \subseteq I?$ Is that because it's trivial? I need to stay within the rings too. – user718615 Oct 30 '19 at 20:11
• Also, I don't get why the degree of the zero polynomial is undefined or "$-\infty$" – user718615 Oct 30 '19 at 20:14
• @user63710 To better understand how to assemble definitions into proofs we should strive to develop a more conceptual view of the proof. For one such view see here, which includes a generalization to any PID (Dedekind-Hasse criterion). – Bill Dubuque Oct 30 '19 at 20:14