Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the importance of integral domains? In abstract algebra Thanks for your help

share|improve this question
I guess that you will not be able to find a right answer. There is no reason to be more or less important. You can obtain interesting facts using integral domains or not. –  Sigur Aug 17 '12 at 1:11
Sidestepping the subjective question of importance, I think the main usefulness of IDness is in left and right cancellation. –  anon Aug 17 '12 at 1:24
From an algorithmic point of view, cancellation law is a very nice property if you're writing an algorithm to perform some computations. But that's in no way a mathematical motivation of course. –  user2468 Aug 17 '12 at 1:57

3 Answers 3

up vote 8 down vote accepted

Definition: An Integral domain is a commutative ring( with unity) $(R,+,\cdot)$ for which the following property holds

If $a,b \in R$ such that $a,b \neq 0$, then $ab\neq 0$.

To see some motivation behind the definition of an Integral domain, notice that when we were little kids, while solving polynomial equations in $\mathbb Z$, after factoring a polynomial, we make the jump from



$$x=0 \text{ or } x=2 \text{ or } x=3$$

This is a convenient property to have, since it almost always makes it easy to find ALL solutions of a polynomial equation.And this property holds for all elements if and only if the ring is an integral domain. This is quite easy to verify.

Another motivation is the following question:

Given a ring $R$, when does there exist a field $F$ and an injective ring homomorphism $\phi:R \to F$?

Informally, we are asking when a given ring $R$ can be described as a subring of a field. The answer turns out that this is possible if and only if $R$ is an integral domain.

One direction of this is easy. Suppose that $R$ can be embedded into a field as above. For any two elements $a,b\neq 0 \in R$, if $ab=0$, then $\phi(ab)=0$ too. Since $\phi$ is a homomorphism, this means that $\phi(a)\phi(b)=0$, Since $\phi$ is injective, neither $\phi(a)$ or $\phi(b)$(Both elements of $F$) are zero. Since $F$ is a field, this means they must be invertible. Multiplying on the left by $(\phi(a))^{-1}$, we see that $\phi(b)=0$, a contradiction. So, $ab \neq 0$. So, $R$ is an integral domain.

For the other direction, see Field of Fractions. This is a construction which not only constructs a field containing our Integral Domain, but also constructs the smallest such one. For example, the Field of Fractions of $\mathbb Z$ is $\mathbb Q$, the field of Rational numbers. This might clarify which the construction is called the field of fractions, since $\mathbb Q$ is what we usual think of as fractions.

Thus, the notion of an abstract Integral domain is closely modeled after the Integers.

share|improve this answer
Moreover, it's a commutative ring, and often assumed to have multiplicative identity. –  Cameron Buie Aug 17 '12 at 1:57
@CameronBuie: Dear Cameron, thank you. Please feel free to edit if you see any mistakes. I am a beginner to this material, afterall.:) –  Galois Group Aug 17 '12 at 1:58
Quite alright. I'm not sure you'd be notified of edits to your post, though, and this way, you also learn something else about the subject. ^_^ –  Cameron Buie Aug 17 '12 at 2:01

@Cameron Buie is correct, if a bit brief. Integral domains eliminate two "undesirable" properties from a general ring: non-commutativity, and zero divisors. I think the latter property is more important, because if we eliminate zero divisors, then we gain the cancellation rule: if $ca = cb$, then $a = b$. We can't say this if we have zero divisors: i.e. $a,b$ are zero divisors if $ab=0$ but $a,b\ne0$. Zero divisors occur in non-trivial situations: e.g. matrix rings, modulo integer arithmetic for non-prime modulus, etc. Integral domains then go on to have other nice properties (e.g. a quotient field), but on an intuitive level, commutativity and cancellation are very nice properties on their own.

share|improve this answer

It is a very useful abstraction. It rests on a huge field of examples. One thing that is interesting is that if $R$ is an i.d., then $R[X]$is an i.d. too.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.