What is the importance of integral domains?

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

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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

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

$$x(x-2)(x-3)=0$$

$$\text{to}$$

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

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.

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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.:) – Ravi 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.

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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.

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