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 $N(r)<N(b),$ 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 $r_1<a,$ which means that $r_1$ must be zero and thus $I=\langle a\rangle.$ I think I'm missing something here.

  • $\begingroup$ It makes sense for you to compare your proof with the standard proof first, see for example here, and then improve your own proof. $\endgroup$ – Dietrich Burde Oct 30 '19 at 19:10
  • $\begingroup$ @DietrichBurde thanks I really need to improve my abstract algebra skills. I'm new to the field. $\endgroup$ – user718615 Oct 30 '19 at 19:11
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    $\begingroup$ 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 $N(.)$. $\endgroup$ – reuns Oct 30 '19 at 19:14
  • $\begingroup$ @reuns why isn't it $|a-bc| < |b|?$ I got that bc $a=bc+r$ and $r<|b|$ $\endgroup$ – user718615 Oct 30 '19 at 20:05
  • $\begingroup$ sure I meant $|a-bc|<|b|$ $\endgroup$ – 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 $x=aq\in\langle a\rangle$.

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  • $\begingroup$ thanks for telling me the important things that I got wrong! $\endgroup$ – user718615 Oct 30 '19 at 20:06
  • $\begingroup$ 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. $\endgroup$ – user718615 Oct 30 '19 at 20:11
  • $\begingroup$ Also, I don't get why the degree of the zero polynomial is undefined or "$-\infty$" $\endgroup$ – user718615 Oct 30 '19 at 20:14
  • $\begingroup$ @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). $\endgroup$ – Bill Dubuque Oct 30 '19 at 20:14
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    $\begingroup$ @user63710: Please, please, please, do not ask questions about other posts here. Ask them in the relevant post! $\endgroup$ – Arturo Magidin Oct 30 '19 at 21:13

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