Is it true that if a ring is not a UFD then it's not a Euclidean Domain?

I have a ring $R=\mathbb{Z}[\sqrt{-d}]=\{ a+b\sqrt{-d} \mid a,b \in \mathbb{Z} \}$ where $d$ is a square free integer. I want to show that it's not a Euclidean domain.

I have read from Why is $\mathbb{Z}[\sqrt{−n}]$ not a UFD? and it makes sense to me. But is there a way to show that it's not a Euclidean domain without using UFD?

EDITED: Thanks @anon for the comment, that way seems very straight forward and I'm sure it is easier than the one I need to do it with. Our professor wanted us to prove it the following way:

First prove that $\delta(z)>0$ if $z\in R$ is not a unit, then prove that there exists a non-unit element $x\in R$ such that $\delta(x) \le \delta(z)$ for every non-unit element $z \in R$, then show that a remainder from the division by $x$ must be a unit in $R$, and that for every $w \in R$ at least one of the three elements $w, w+1$, and $w-1$ must be divisible by $x$ in $R$. Lastly, show that one can pick $w \in \mathbb{Z} \subset R$ such that none of the elements $w$, $w+1$, and $w-1$ is divisible by $x$ in $R$.

And I have no idea how I need to proceed from this.

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    $\begingroup$ Every ED is a UFD. Thus not-UFD implies not-ED. This is called the "contrapositive," it is a very nifty concept from logic to be familiar with. I am curious as to why you would want to prove non-EDness without proving non-UFDness; is it clear there will be a better way? $\endgroup$ – anon Dec 15 '13 at 7:58
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    $\begingroup$ Here's a geometric way: math.stackexchange.com/questions/318543/…? $\endgroup$ – tc1729 Dec 15 '13 at 17:11

The method suggested by your professor is sometimes called the universal side divisor criterion for showing that a ring in non-Euclidean. The basic idea is quite simple. To prove that a domain is non-Euclidean it suffices to show that it lacks some property possessed by all Euclidean domain. Here this property is that if $x$ is a nonunit of minimal Euclidean value, then, modulo $x$, every element has remainder $0$ or a unit (else the remainder is a nonunit having Euclidean value smaller than $x$, contra hypothesis).

Based on your hint, in your case it seems that the only units of $R$ are $\pm 1$, so every element $w$ has remainder $\,0\,$ or $\,\pm1,\,$ i.e. $x$ divides $w$ or $w\mp1$. It remains to show that there is some $w$ for which this fails, thus completing the proof that $R$ is not Euclidean.

If you google "universal side divisor" you will find some insightful expositions, including worked examples. For example see Keith Conrad's Remarks about Euclidean domains. See also my answer here which outlines a proof that $\rm\: \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\ $ is a non-Euclidean PID - based on sketch by Hendrik W. Lenstra.

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  • $\begingroup$ Thanks for the great helps! I have two questions. How will I show that $\delta(z)>0$ if $z$ is not a unit, and for the last step, do I just pick a $w$ that doesn't work then I'll be done? Thanks! Bill Bubuque & @YACP $\endgroup$ – PandaMan Dec 17 '13 at 4:59
  • $\begingroup$ @Peter Which specific quadratic ring are you working with? $\endgroup$ – Gone Dec 17 '13 at 5:09
  • $\begingroup$ $R=\mathbb{Z}[\sqrt{-d}]=\{ a+b\sqrt{-d} \mid a,b \in \mathbb{Z} \} $ @Bill $\endgroup$ – PandaMan Dec 17 '13 at 5:15
  • $\begingroup$ For my first question, is it just because $(a+b\sqrt{-d})(a-b\sqrt{-d})=(a^2+b^2d)$ which is always positive? $\endgroup$ – PandaMan Dec 17 '13 at 5:19
  • $\begingroup$ @Peter Generally, Euclidean $\Rightarrow$ UFD, so, conrapositively, non-UFD $\Rightarrow$ non-Euclidean. But usually the proof that such criterion holds cannot be given generally. It requires invoking specific properties of the ring. As such, you should have be given a specific value of $d,$ such as $d = 19$ in the linked posts. $\endgroup$ – Gone Dec 17 '13 at 5:22

Every Euclidean domain is a UFD, so if a ring is not a UFD, then it is not a Euclidean domain. Note that $\mathbb{Z}[\sqrt{-d}]$ is a Euclidean domain for certain values of $d$. See Wikipedia for more details.

Edit I see you've edited your question. Presumably you want to assume $d > 2$ then.

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  • $\begingroup$ thanks, I misread it backwards, just UFD was a subset of ED. $\endgroup$ – PandaMan Dec 17 '13 at 5:00

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