Ring Theory/Integral domains and Fields

Definition 1: A non zero element 'a' of a commutative ring R is called a zero divisor if there exists some non zero element b in R such that ab=0.

For example, in the ring of 2-by-2 matrices, the matrix

${\displaystyle {\begin{pmatrix}1&1\\2&2\end{pmatrix}}}$

is a zero divisor because

${\displaystyle {\begin{pmatrix}1&1\\2&2\end{pmatrix}}\cdot {\begin{pmatrix}1&1\\-1&-1\end{pmatrix}}={\begin{pmatrix}-2&1\\-2&1\end{pmatrix}}\cdot {\begin{pmatrix}1&1\\2&2\end{pmatrix}}={\begin{pmatrix}0&0\\0&0\end{pmatrix}}.}$

Definition 2: A commutative ring is called an integral domain, if it has no zero divisors. Equivalently, a commutative ring is called an integral domain if ${\displaystyle ab=0\Rightarrow a=0\ or\ b=0}$ or in other words ${\displaystyle a\neq 0,b\neq 0\Rightarrow ab\neq 0\forall a,b\in R}$.

For example ${\displaystyle \mathbb {Z} ,\mathbb {Q} ,\mathbb {R} ,\mathbb {C} }$ are all integral domains. ${\displaystyle \mathbb {Z} _{6}}$ is not an integral domain (2.3=0 here) but ${\displaystyle \mathbb {Z} _{5}}$ is.

Definition 3: A ring (R,+,.) is called a division ring if it forms a group with respect to the operation '.'. If that group is abelian then the ring is called a field. We will emphasize the properties of a field again:

A field F is a set together with two operations, usually called addition and multiplication, and denoted by + and ·, respectively, such that the following axioms hold:

• Closure of F under addition and multiplication

For all a, b in F, both a + b and a · b are in F (or more formally, + and · are binary operations on F).

• Associativity of addition and multiplication

For all a, b, and c in F, the following equalities hold: a + (b + c) = (a + b) + c and a · (b · c) = (a · b) · c.

• Commutativity of addition and multiplication

For all a and b in F, the following equalities hold: a + b = b + a and a · b = b · a.

There exists an element of F, called the additive identity element and denoted by 0, such that for all a in F, a + 0 = a. Likewise, there is an element, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a. For technical reasons, the additive identity and the multiplicative identity are required to be distinct.

For every a in F, there exists an element −a in F, such that a + (−a) = 0. Similarly, for any a in F other than 0, there exists an element a−1 in F, such that a · a−1 = 1. (The elements a + (−b) and a · b−1 are also denoted a − b and a/b, respectively.) In other words, subtraction and division operations exist.

• Distributivity of multiplication over addition

For all a, b and c in F, the following equality holds: a · (b + c) = (a · b) + (a · c).

Clearly every field is a division ring. The easiest examples of fields are ${\displaystyle \mathbb {Q} ,\mathbb {R} }$ and ${\displaystyle \mathbb {C} }$. A division ring which is not a field is the field of quaternions, described as follows:

Consider ${\displaystyle \mathbb {R} ^{4}}$. Let the symbol 1 stand for (1,0,0,0); i for (0,1,0,0); j for (0,0,1,0) and k for (0,0,0,1). Clearly every element of ${\displaystyle \mathbb {R} ^{4}}$ can be represented as ${\displaystyle \alpha _{0}(1)+\alpha _{1}(i)+\alpha _{2}(j)+\alpha _{3}(k)}$ where ${\displaystyle \alpha _{i}}$ is some real number. We endow addition and multiplication on ${\displaystyle \mathbb {R} ^{4}}$ according to the following rules: Addition of two elements ${\displaystyle \alpha _{0}(1)+\alpha _{1}(i)+\alpha _{2}(j)+\alpha _{3}(k)}$ and ${\displaystyle \beta _{0}(1)+\beta _{1}(i)+\beta _{2}(j)+\beta _{3}(k)}$ is simply ${\displaystyle (\alpha _{0}+\beta _{0})(1)+(\alpha _{1}+\beta _{1})(i)+(\alpha _{2}+\beta _{2})(j)+(\alpha _{3}+\beta _{3})(k)}$. For multiplication note that if we impose the following rules:

${\displaystyle i^{2}=j^{2}=k^{2}=ijk=-1,}$

then these determine all the possible products of i, j, and k.

For example, since

${\displaystyle -1=ijk,}$

right-multiplying both sides by k gives

{\displaystyle {\begin{aligned}-k&=ijkk,\\-k&=ij(-1),\\k&=ij.\end{aligned}}}

All the other possible products can be determined by similar methods, and this gives the following table:

{\displaystyle {\begin{alignedat}{2}ij&=k,&\qquad ji&=-k,\\jk&=i,&kj&=-i,\\ki&=j,&ik&=-j.\end{alignedat}}}

For two elements ${\displaystyle \alpha _{0}(1)+\alpha _{1}(i)+\alpha _{2}(j)+\alpha _{3}(k)}$ and ${\displaystyle \beta _{0}(1)+\beta _{1}(i)+\beta _{2}(j)+\beta _{3}(k)}$, their product is determined by the products of the i,j,k's according to the above rules and the distributive law. This gives the following expression:

${\displaystyle \alpha _{0}\beta _{0}-\alpha _{1}\beta _{1}-\alpha _{2}\beta _{2}-\alpha _{3}\beta _{3})(1)+(\alpha _{0}\beta _{1}+\alpha _{1}\beta _{0}+\alpha _{2}\beta _{3}-\alpha _{3}\beta _{2})(i)+(\alpha _{0}\beta _{2}-\alpha _{1}\beta _{3}+\alpha _{2}\beta _{0}+\alpha _{3}\beta _{1})(j)+(\alpha _{0}\beta _{3}-\alpha _{1}\beta _{2}-\alpha _{2}\beta _{1}+\alpha _{3}\beta _{0})(k)}$

It is left to the reader to verify that the thus obtained algebraic structure is indeed a division ring.

Basic Theorems on Integral domains and fields

Theorem 1.11: Let R be a commutative ring. Then R is an integral domain if and only if ${\displaystyle ab=ac\Rightarrow b=c}$ where ${\displaystyle a,b,c\in R,a\neq 0}$.

Proof: ${\displaystyle \Rightarrow }$: Clearly ab=ac implies a(b-c)=0. As a is non zero and R is an integral domain so b-c=0 or b=c.

${\displaystyle \Leftarrow }$: Suppose that for some nonzero a we have ab=0. But then ab=a0 and by our hypothesis b=0.${\displaystyle \Box }$.

Remark: Basically the above theorem means that integral domains are the rings where cancellation laws hold. In rings where cancellation laws do not hold we are bound to have some zero divisors.

Theorem 1.12: Every field is an integral domain.

Proof: Let R be any field. Let ab=ac, where ${\displaystyle a,b,c\in R}$ and ${\displaystyle a\neq 0}$. Then as ${\displaystyle a^{-1}}$ exists so multiplying it on both sides of ab=ac we have b=c, i.e. cancellation laws hold. By the previous theorem R is an integral domain.${\displaystyle \Box }$

Remark: The converse of the above result may not be true as is evident from ${\displaystyle \mathbb {Z} }$.

Theorem 1.13: Every finite integral domain is a field.

Proof: Let R be a finite integral domain and let ${\displaystyle x\in R}$ where ${\displaystyle x\neq 0,1}$. It suffices to show that x is a unit. Now the list 1,x,x2,x3... can't go on forever as R is finite. Suppose, without losing generality that for some i<j, xi=xj. Then xi-xj=0 and since i<j, so xj-i is a legitimate member of R (in fact so is xj-i-1). We have xi(1-xj-i)=xi-xj=0. As x is non zero and R is an integral domain so xi is non zero. But then 1-xj-i=0 or xj-i=1. It follows that as xj-i-1x=1. Hence x is a unit with inverse xj-i-1.${\displaystyle \Box }$

Corollary: The ring ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ is a field iff p is prime.

Proof: ${\displaystyle \Rightarrow }$: We will denote elements of ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ by numbers 0,1,...p-1. Now suppose p was composite and p=ab where 1<a,b<p. Now ab=0 in ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ although a,b are themselves nonzero. This contradicts the fact that ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ is an integral domain.

${\displaystyle \Leftarrow }$ Suppose p is prime. It suffices to show that ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ is an integral domain. Let a,b be nonzero elements of ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ such that ab=0 there. But then p|ab and as p is prime so p|a or p|b. That's just another way of saying that a=0 or b=0 in ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ and so ${\displaystyle {\frac {\mathbb {Z} }{p\mathbb {Z} }}}$ is an integral domain.${\displaystyle \Box }$

Theorem 1.14: Let R be a ring such that the equation ax=b has a solution for all ${\displaystyle a(\neq 0)\in R}$ and for all ${\displaystyle b\in R}$. Then R is a division ring.

Proof: We first show that R has no zero divisors followed by the fact that it has a unity. Let ab=0 where a,b are non zero. Now abx=0 for each x in R. Ler r be any element of R. Now by the hypothesis there exists an x such that bx=r. Using this x we see that ar=0 for any r in R. Now consider ax=a. Clearly there is a c such that ac=a. But ar=0 implies that ac=0=a. This contradicts the fact that a was chosen to be nonzero. So R has no zero divisors. Now let e be the solution of ax=a. Obviously e is nonzero. Then ae=a and a(e-e2)=ae-ae2=a-ae=0 and so e=e2. Then for any x (xe-x)e=xe-xe=0 and as e is nonzero so xe=x following which R has unity. (The fact that ex=x is similarly proved.)

Now if a is non zero, then ax=e has a solution a-1. Also (a-1a-e)a-1=a-1e-ea-1=0 and so as a-1 is non zero we have a-1a-e=0 or a-1a=e. Then aa-1=a-1a=e and so a is a unit. Similarly all non zero elements are units.${\displaystyle \Box }$