# Abstract Nonsense

## 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.