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I recently started learning about rings and some of their elementary characteristics / basic properties. One of the concepts that sort of caught me off guard was the statement that infinite ordered integral domains are not necessarily fields. I thought about it and saw that $\mathbb Z$ was a concrete example of this because other than the elements $1, -1$, no other elements have multiplicative inverses.

Integral domains are defined as: a commutative ring with unity having the cancellation property $\iff$ commutative ring with unity having no divisors of $0$

Fields are defined as: a commutative ring with unity in which every nonzero element is invertible

From prior learning about groups, the proof that groups exhibit the cancellation product employed a strategy that invoked inverse elements. (i.e. $ax=bx \implies axx^{-1}=bxx^{-1} \implies a=b$)

If a particular integral domain is not a field (and therefore exhibits the cancellation property but not all elements have multiplicative inverses), does that mean that the cancellation property of some integral domains is fundamentally different than the cancellation property of a group?

I ask this because the proof strategy for demonstrating that such an integral domain exhibits the cancellation property must be fundamentally different from the strategy that is used in the group-proof (because invertible elements are generally absent).

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    $\begingroup$ An integral domain (or general ring) $(R,+,\cdot)$ is a group under $+$ addition. The group $(R,+)$ exhibits the cancellation property with respect to $+$, but, as you pointed out, not necessarily with respect to multiplication $\cdot$. $\endgroup$
    – mi.f.zh
    Commented Jan 11, 2020 at 12:25
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    $\begingroup$ It might be helpful to look at the wikipedia page if you haven't already - note that it is a property held by any operation, addition or multiplication (or whatever else). $\endgroup$
    – mi.f.zh
    Commented Jan 11, 2020 at 12:26
  • $\begingroup$ @nhmwhhxx I'm a little confused then. In my book, I am told that an integral domain has "the cancellation property". It does not mention anything about "with respect to a particular operation". $\endgroup$
    – S.C.
    Commented Jan 11, 2020 at 12:33
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    $\begingroup$ It's a bit unfortunate with the terminology, but implicitly I'm sure the author means 'with respect to multiplication'. Again, from the wikipedia page, the cancellation property is defined for a groupoid with an abstract binary operation - in a ring, we have two distinctly defined operations, addition and multiplication. Hence, to specify the cancellation property one must (implicitly in the author's case) specify the operation. $\endgroup$
    – mi.f.zh
    Commented Jan 11, 2020 at 12:41

2 Answers 2

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If you look at the case of the integers $\mathbb Z$, you can prove that it has the cancellation property by using the fact that it has no zero divisors. However, many people would take an approach that is technically much more difficult by arguing that $\mathbb Z$ can be embedded (as a ring) in the field of rational numbers $\mathbb Q$, in which the cancellation property holds because of the existence of inverses. This other argument can also be extended to arbitrary integral domains, by considering the field of fractions of the integral domain.

Whether this makes the cancellation property fundamentally the same for integral domains and fields, I'll leave for you to judge. I think the important point is that, even if such constructions are possible, they are by no means needed to prove the cancellation property.

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  • $\begingroup$ Just for clarification, given the comments that @nhmwhhxx made above, when my book says "integral domains exhibit the cancellation property"...the implicit interpretation is that "integral domains exhibit the cancellation property FOR THE MULTIPLICATION operation". The reason it is implicit is because it is already understood that, by definition, the ring's addition operation is a group...and therefore a ring always satisfies the cancellation property for the addition operation. Is that correct? $\endgroup$
    – S.C.
    Commented Jan 11, 2020 at 12:42
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    $\begingroup$ Yes, that is correct. In the same way, when one says a ring is commutative, it is meant that the multiplication is commutative, since the addition is already commutative by definition. $\endgroup$
    – Krup'a
    Commented Jan 11, 2020 at 13:10
  • $\begingroup$ @Krup'a: I don't understand this answer. What is "technically much more difficult" about embedding $\mathbb{Z}$ into the rational numbers? $\endgroup$ Commented Dec 9, 2020 at 9:01
  • $\begingroup$ If you haven't constructed the rational numbers yet, you would first have to construct them out of the integers, which is much more complicated (more than two lines of work) than doing the standard proof using the fact that $\mathbb Z$ has no zero divisors (two lines of work). Of course, if you know already the rationals exist it's not complicated at all. In contrast, for an arbitrary integral domain, you generally don't "know' the field of fractions. You could construct it to prove the cancellation property, but that's conceptually a big detour for something that's actually very easy. $\endgroup$
    – Krup'a
    Commented Dec 12, 2020 at 14:51
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In order to prove that, in a commutative ring $(R,+\times)$, the cancellation property holds, you cannot assume that every non-zero element has an inverse; you are not assuming that $(R\setminus\{0\},\times)$ is a group (if it was, your commutative ring would be a field.

For instance, $\mathbb Q[x]$ is an integral domain in checking this means, in particular, that you should check that$$P(x),Q(x)\in\mathbb Q[x]\setminus\{0\}\implies P(x)Q(x)\neq0.$$

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