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It may sound like a basic question, but I am finding it hard to prove using the definition of Euclidean domain.

So far I have thought that there should be an element $a \in R$ such that $d(a)$ is minimum. This will imply that for any $x \in R$, $x = pa + r$ where $r=0$ or $d(r) < d(a)$. But as $a$ is chosen such that $d(a)$ is minimum, $r$ must be $0$.

So, $a$ should divide every element of $R$. I am stuck here.

Definition provided in the book: An integral domain R is called a Euclidean domain if for all $a \in R$, $a \neq 0$ there is defined a non -ve integer $d(a)$ s.t.,

  1. for all $a,b \in R$, $a \neq 0$, $b \neq 0$, $d(a) \leq d(ab)$
  2. for all $a,b \in R$, $a \neq 0$, $b \neq 0$, $\exists$ $t$ and $r$ in $R$ s.t., $a = tb + r$, where either $r=0$ or $d(r) \lt d(d)$.

Definition of integral domain in the book: "A commutative ring $R$ is called an Integral domain if $ab=0 \implies$ either $a=0$ or $b=0$".

Prince Kumar
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    finding it hard to prove using the definition of Euclidean domain What definition are you using? A Euclidean domain is a commutative ring, which has a multiplicative unit by (the usual) definition. – dxiv Aug 09 '17 at 06:54
  • Throughout the book any general ring is assumed without unity. In the definition, existence of unity is not explicitly mentioned. It is just defined with a function $d$ on set of non-zero elements with the properties: (1) $d(x) \leq d(xy)$ and (2) for every $a,b \in R$, there exists $q,r \in R$ such that $a = qb +r$. – Prince Kumar Aug 09 '17 at 07:02
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    Is there a definition of "domain" in your book? A Euclidean domain is of course a domain with some extra properties. – drhab Aug 09 '17 at 07:09
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    @PrinceKumar You'll help your question get better answers if you mentioned what book that is (maybe someone is familiar), and fully quote the definitions (those (1) and (2) must certainly say more than you wrote, since that is not nearly enough to make much sense of them). – dxiv Aug 09 '17 at 07:14
  • @dxiv Ring was additionally assumed to be an integral domain. Edited to included exact definition provided in the book. – Prince Kumar Aug 09 '17 at 07:29
  • Part of the definition of integral domain is existence of unity. – Gerry Myerson Aug 09 '17 at 07:29
  • @GerryMyerson Throughout the book it is assumed that every ring mentioned is without unity unless otherwise stated. The definition of an integral domain as provided in the book: "A commutative ring $R$ is called an Integral domain if $ab=0 \implies$ either $a=0$ or $b=0$". – Prince Kumar Aug 09 '17 at 07:34
  • That would make the even integers an integral domain. I've never seen anyone use the term that way. – Gerry Myerson Aug 09 '17 at 07:35
  • @GerryMyerson You can go through the first comment on this question: https://math.stackexchange.com/questions/420325/why-ring-with-only-even-numbers-is-not-an-integral-domain – Prince Kumar Aug 09 '17 at 07:38
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    A general ring might be without unit in your book. What about a commutative ring? – Arthur Aug 09 '17 at 08:04
  • Why do you think the given definitions imply the existence of a unit? Will the even integers not precisely give you an example of a non-unital ring satisfying these conditions? – Tobias Kildetoft Aug 09 '17 at 08:09
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    Actually, the even integers do not satisfy the requirements, and working through how they fail seems to give a good idea of how to show that these conditions imply the existence of a unity (or at least something close to it). – Tobias Kildetoft Aug 09 '17 at 08:38
  • @TobiasKildetoft I am not sure how to compare even integers with the general case. First I thought about defining $d(2n) = n$. But this does not satisfy the second condition for $a=6$, $b=4$. But this does not prove that there can not be any such function $d$. Any thoughts? – Prince Kumar Aug 09 '17 at 08:59
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    Consider the obvious function (being the same as the one from the integers). Now let $a = b$ in condition 2 and see what fails. – Tobias Kildetoft Aug 09 '17 at 09:01
  • That helped. I used the condition that $a$ divides itself to arrive that there is an element $t$ such that $a = at$. So $ab = atb \implies a(b-tb)=0 \implies b = tb$. So $t$ must be the identity of R. – Prince Kumar Aug 09 '17 at 09:36
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    @PrinceKumar For the purposes of learning the basic theory of factorization (with PID's, UFD's and ED's) it doesn't make a lot of sense to start with rings lacking identity. For one thing, if you don't have an identity, you can't talk about units, and can't talk about associate elements. – rschwieb Aug 09 '17 at 13:14
  • Although your book doesn't work with unitary rings in general, I'm pretty sure the author is working with integral domains with unity when he develops the theory of factorization on integral domains. – Xam Aug 09 '17 at 15:50
  • I am not sure why people are downvoting this question. Is it because they prefer to answer questions about rings with unity only? – Prince Kumar Aug 12 '17 at 15:19

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As $a$ divides every element of $R$, it divides itself. So $a = ta$ for some $t \in R$. So for every $b \in R$, $ab = tab \implies a(b-tb)=0 \implies b=tb$. So $t$ is the identity of $R$.

Prince Kumar
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  • How do you guarantee that there is $a$ such that $d(a)$ is minimum? – Xam Aug 09 '17 at 15:53
  • @Xam $d$ takes only non-negative integral values. So, the set ${d(r) | r \in R }$ has non-negative integral values. So, it must have a minimum. Set $a$ to be a $r$ (there can be multiple such $r$) which attains this minimum. – Prince Kumar Aug 12 '17 at 15:17