Abstract Nonsense

Crushing one theorem at a time

Integral Domains


Point of Post: In this post we discuss the notion of integral domains.

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Motivation

Intuitively, integral domains are just commutative rings where you can’t multiply non-zero elements of the ring and get back zero. In this post we’ll define this more rigorously and prove that every finite integral domain is a field.

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Integral Domain

Let R be a unital commutative ring with 1\ne 0. We say that R is an integral domain if it has no zero divisors. In other words, R has the property that ab=0 implies that a=0 or b=0. We now give some different characterizations of this:

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Theorem: Let R be a commutative unital ring with 1\ne 0. Then, R is an integral domain if and only if ab=ac implies b =c for every a\ne 0,b,c\in R and ba=ca implies b =c for a\ne0,b,c\in R

Proof: Suppose first that R is an integral domain. The fact that ab=ac implies that ab-ac=0 or that a(b-c)=0. But, since R is an integral domain we know that a=0 or b-c=0 and since a\ne0 we have that b-c=0 or that b =c. The fact that ba=ca implies that b=c is done similarly.

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Conversely, suppose that ab=ac implies b =c for every a\ne0,b,c\in R. Suppose further that ab=0 for a,b\in R. If a\ne 0 we have that ab=0=a0 and so b=0. Similarly, if b\ne0 then ab=0=0b and so a=0. Either way, we’re done$. \blacksquare

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If a ring has the property that ab=ac implies b =c for non-zero a we call the ring left cancellative. If the analogous property holds on the right we call the ring right cancellative. If both of these hold we call the ring just cancellative. Thus, in light of this theorem we see that an integral domain is a unital commuative cancellative ring with 1\ne 0.

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Note though the clear fact that a ring R is cancellative if and only if the maps R\to R:x\mapsto rx and R\to R:x\mapsto xr are injective for every r\in R.

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With this in mind we can now prove that every finite integral domain is a field.

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Theorem: Every finite integral domain R is a field.

Proof: Since R is already a commutative unital ring with 1\ne 0 it suffices to show that it is a divison ring. To do this we let r\in R be arbitrary and consider the map f_r:R\to R:x\mapsto rx. By the above remark we know that f_r is injective, and since R is finite we may appeal to a common set-theoretic fact to conclude that f_r is surjective. Thus, there exists s\in R such that 1=f_r(s)=rs and since R is commutative we know that sr=rs=1. Thus, since r was arbitrary the conclusion follows. \blacksquare

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So what’s the quintessential example of a commutative unital ring with 1\ne 0 which is not an integral domain? Consider C(\mathbb{R}), the set of all continuous functions \mathbb{R}\to\mathbb{R}. This is evidently a ring with the usual addition and multiplication of functions. Moreover, it’s clear that it’s commutative and unital with 1 being the constant function 1(x)=1. That said, the functions f(x)=x\mathbf{1}_{[0,\infty)} (where \mathbf{1}_{[0,\infty)} is the indicator function on [0,\infty)) and g(x)=x\mathbf{1}_{(-\infty,0]} are evidently non-zero elements of C(\mathbb{R}) yet it’s easy to see that fg=0.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

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June 15, 2011 - Posted by | Algebra, Ring Theory | , , , ,

2 Comments »

  1. […] there is a meaningful equality for in terms of , but this requires that we assume that is an integral domain. […]

    Pingback by Polynomial Rings (Pt. II) « Abstract Nonsense | July 19, 2011 | Reply

  2. […] like in the case of domains in , is a ring for any Riemann surface, but more specifically, is an integral domain (this is where the term ‘domain’ in integral domain came from, by the […]

    Pingback by Holomorphic Maps and Riemann Surfaces (Pt. IV) « Abstract Nonsense | October 4, 2012 | Reply


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