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Background:

Definition: An integral domain $R$ is a Euclidean domain if there is a function $\delta$ from the nonzero elements of $R$ to the nonnegative integers with these properties;

(i) If $a$ and $b$ are nonzero elements of $R$, then $\delta(a)\leq \delta(ab)$.

(ii) If $a,b\in R$ and $b\neq 0_R$, then there exist $q,r\in R$ such that $a=bq+r$ and either $r=0_R$ or $\delta(r) < \delta(b)$.

Exercise 1: Let $R$ be a Euclidean ring with Euclidean norm $\delta$. Let $a,b\in R\setminus\{0\}$ and let $q,r\in R$ such that $a=bq+r$ with $r=0$ or $\delta(r)<\delta(b)$. Prove that $r$ and $q$ are unique if and only if $\delta(a+b)\le\max\{\delta(a),\delta(b)\}$.

Proof: Let $t,r$ be uniquely determined and suppose

$\delta(a+b)> \text{max}\{\delta(a), \delta(b)\}$ for some $a,b$ (non zero) in $R$.

Now $b=0\cdot(a+b)+b=1\cdot(a+b)-a$

Also $\delta(-a)=\delta(a)<\delta(a+b)$

and $\delta(b)<\delta(a+b)$

Thus for $b, 1\in R$, there exists a $t=0$, $r=b$ or $t_1=1$, $r_1=-a$ such that $b=t$, $1+r$, $b=t_1\cdot 1 +r_1$ where $r\neq r_1$ (as $a+b\neq 0) t\neq t_1$, a contradiction to uniqueness.

Hence $\delta(a+b)\leq \text{max}(\delta(a),\delta(b))$.

Questions:

In the above proof, where it says: "Now $b=0\cdot(a+b)+b=1\cdot(a+b)-a$ Also $\delta(-a)=\delta(a)<\delta(a+b)$", why is it necessary for the author to have $b=0\cdot(a+b)+b=1\cdot(a+b)-a$ and also where did $\delta(a)$ came from? I understand that in the division algorithm from Definition above, that for $b=(a+b)$ and $r=-a$, and $r=b$, $q=0,1$ at this point of the proof. I just don't understand the motivation behind its logic.

Thank you in advance

Arturo Magidin
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Seth
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    Note: a few weeks ago, OP asked a prior question on the same exercise. – Bill Dubuque Jan 18 '24 at 00:19
  • @BillDubuque I am asking a different question for the same exercsie. I am asking for understanding the proof. I reread the proof and there was something that I was confused about and I am asking for clarification from the community. – Seth Jan 18 '24 at 00:22
  • The proof shows that if $a,b$ are both smaller than their sum then $,-a,b,$ are both possible remainders left on dividing $b$ by $a+b.,$ Since $,-a\neq b,$ this shows that the division does not have a unique remainder. – Bill Dubuque Jan 18 '24 at 00:26
  • @BillDubuque I asked a similar question here. It seem the technique: $b=0\cdot(a+b)+b=1\cdot(a+b)-a$ is very common amongst these type of questions. But why do we have to write $b$ in two different ways. I am just not understanding the logic behind it. – Seth Jan 18 '24 at 00:39
  • @BillDubuque ah kk. Thank you for the clarification. I did not know that is what the author is using. – Seth Jan 18 '24 at 00:46
  • @BillDubuque if it is not too much trouble, can I ask you to take a loook at the following post1 post2, they are related but they are asking two different things. It h as to do with quotient ring of gaussian integers. Also I will delete my other post that is similar to this one. Thank you in advance. – Seth Jan 18 '24 at 00:50
  • e.g. for $,2\div (2!+!1),,$ we have $,2 = 0(2!+!1)\color{#0a0}{+2} = 1(2!+!1)\color{#c00}{-1}.,$ Both $,r = \color{#0a0}{+2},,\color{#c00}{-1}$ satisfy $,|r| < |2+1|,$ so both $,r,$ are valid remainders, so remainders are not unique in $\Bbb Z$ with norm $,\delta(n) = |n|.\ $ It's an obvious generalization familiar from balanced remainder systems in $\Bbb Z,,$ – Bill Dubuque Jan 18 '24 at 00:52
  • e.g. we prefer $,{-}1$ (vs. $10)$ for $,10\bmod 11,,$ when casting out elevens. That we have these $\rm\color{#c00}{two}$ possible smaller magnitude ($\delta$-size) choices is what destroys remainder $\rm\color{#c00}{uniqueness}$ in $\Bbb Z.\ \ $ – Bill Dubuque Jan 18 '24 at 00:59
  • @Seth: The statement in the exercise is not quite correct unless one adds that $\delta(a+b)\leq \max(\delta(a),\delta(b)$ of all $a,b\in R\setminus{0}$. – Mittens Jan 18 '24 at 01:14
  • @Mittens ah kk thank you for letting me know. – Seth Jan 18 '24 at 01:20
  • @Seth: Here is a page that might be of interest – Mittens Jan 18 '24 at 01:34
  • @Mittens ah kk thank you for the référence. – Seth Jan 18 '24 at 03:48
  • @Mittens can I trouble you with something very quick. Do you know in which Euclidean domain where uniqueness of $q,r$ or $\delta(a+b)\leq\max{\delta(a),\delta(b)}$ can be satisfied. I try a lot of popular examples from textbooks, but I am not haveing any sucess for the later condition and hence uniqueness of $q,r$ would fail. I am wondering if the exercise is only valid over polynomial rings or say Gaussian integers. I don't know much about Algebraic number theory at the moment. – Seth Jan 19 '24 at 02:39
  • @Seth: I think in the ring of polynomials with $\delta(f)=\operatorname{deg}(f)$ of $f\neq0$. The condition you described in your posting is satisfied. – Mittens Jan 19 '24 at 02:45
  • @Mittens so it is only with univariate polynomial rings? But it won't work for multivariate polynomial rings? – Seth Jan 19 '24 at 05:10

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