Ring Theory/Printable version

Ring Theory

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Ring extensions

Definition (ring extension):

Whenever $S$ is a ring and $R\subset S$ is a subring of $S$, we say that $S$ is a ring extension of $R$ and write $S/R$.

Note that if $S/R$ is a ring extension, then $S[x]/R[x]$ is a ring extension; indeed, the set $S[x]$ is the set of all polynomials with coefficients in $S$, the set $R[x]$ is the set of all polynomials with coefficients in $R$, and $R[x]$ is a subring of $S[x]$.

Proposition (existence of splitting ring):

Let $R$ be a ring, and let $p\in R[x]$ be a polynomial over $R$. Then there exists a ring extension $S/R$ such that in ${\overline {S}}$, $p$ decomposes into linear factors, that is,

$p(x)=(x-\lambda _{1})\cdots (x-\lambda _{n})$ for certain $\lambda _{1},\ldots ,\lambda _{n}\in {\overline {S}}$.

Proof: We prove the theorem by induction on the degree of $p$. Suppose first that $p$ can be decomposed into two polynomials $\Box$

Introduction

The study of rings originated from the theory of polynomial rings and the theory of algebraic integers. Furthermore, the appearance of hypercomplex numbers in the mid-nineteenth century undercut the pre-eminence of fields in mathematical analysis.

Richard Dedekind introduced the concept of a ring.

The term ring (Zahlring) was coined by David Hilbert in the article Die Theorie der algebraischen Zahlkörper, Jahresbericht der Deutschen Mathematiker Vereinigung, Vol. 4, 1897.

The first axiomatic definition of a ring was given by Adolf Fraenkel in an essay in Journal für die reine und angewandte Mathematik (A. L. Crelle), vol. 145, 1914.

In 1921, Emmy Noether gave the first axiomatic foundation of the theory of commutative rings in her monumental paper Ideal Theory in Rings.

Rings

We will start by the definition of a ring.

Definition 1: A ring is a non empty set R together with two binary compositions defined by + and ., and satisfying the following properties hold for any $a,b,c\in R$:

• $a+b\in R$
• $a+b=b+a$
• $a+(b+c)=(a+b)+c$
• There exists an element denoted by $0\in R$ such that $a+0=a$. 0 is called the additive identity or the zero element in R.
• For each $a\in R$, there exists an element $b\in R$ such that $a+b=0$. b is called additive inverse or negative of a and is written as b=-a so that a+(-a)=0.
• $a.b\in R$
• $a.(b.c)=(a.b).c$
• $a.(b+c)=a.b+a.c$ (Left distributive law.)
• $(a+b).c=a.c+b.c$ (Right distributive law.)

We denote a ring by (R,+,.). When the context is clear we just talk about a ring R and assume that the operations + and . are implicit. We will also drop the . in the operation a.b and just say ab.

The first 5 axioms of a ring just mean that (R,+) is an abelian group. The next two mean that (R,.) is a semi group. A ring is called commutative if $a.b=b.a\ \forall a,b\in R$. A ring is called boolean if $x^{2}=x\ \forall x\in R$. A ring R is called a ring with a unit element or unity or identity if $\exists$ an element $e\in R$ such that $ae=ea=a\ \forall a\in R$. Let R be a ring with unit element e. An element $a\in R$ is called invertible, if there exists an element $b\in R$ such that $ab=ba=e$. If n is a positive integer and a an element of a ring R then we define $a^{n}=\underbrace {aa\cdots a} _{n\ times}$ and $na=\underbrace {a+a\cdots +a} _{n\ times}$.

Examples

One of the most important rings is the ring of integers $\mathbb {Z}$ with usual addition and multiplication playing the roles of + and . respectively. It is a commutative ring with identity as 1. The set of even numbers $2\mathbb {Z} :=\{0,\pm 2,\pm 4\cdots \}$ is an example of a ring without identity. Like $\mathbb {Z}$, the sets of rational numbers $\mathbb {Q}$, of real numbers $\mathbb {R}$ and of complex numbers $\mathbb {C}$ are also rings with identity. However $\mathbb {N}$ is not a ring.

The ring of Gaussian integers is given by the set $\mathbb {Z} [i]=\{m+ni:m,n\in \mathbb {Z} \}$ where usual addition and multiplication of complex numbers are the operations. Here i stands (0,1) as is usual in the complex plane.

The set of all n by n matrices with real entries is an example of a non commutative ring with identity, under the usual addition and multiplication of matrices.

The ring of integers modulo n

We now digress slightly to discuss a special kind of an equivalence relation which gives rise to an important class of finite rings.

Let n be a positive integer. Two integers a and b are said to be congruent modulo n, if their difference a − b is an integer multiple of n. If this is the case, it is expressed as:

$a\equiv b{\pmod {n}}.\,$

The above mathematical statement is read: "a is congruent to b modulo n".

For example,

$38\equiv 14{\pmod {12}}\,$

because 38 − 14 = 24, which is a multiple of 12. For positive n and non-negative a and b, congruence of a and b can also be thought of as asserting that these two numbers have the same remainder after dividing by the modulus n. So,

$38\equiv 2{\pmod {12}}\,$

because both numbers, when divided by 12, have the same remainder (2). Equivalently, the fractional parts of doing a full division of each of the numbers by 12 are the same: .1666... (38/12 = 3.166..., 2/12 = .1666...). From the prior definition we also see that their difference, a - b = 36, is a whole number (integer) multiple of 12 ( n = 12, 36/12 = 3).

The same rule holds for negative values of a:

$-3\equiv 2{\pmod {5}}.\,$

The properties that make this relation a congruence relation (respecting addition, subtraction, and multiplication) are the following.

If $a_{1}\equiv b_{1}{\pmod {n}}$ and $a_{2}\equiv b_{2}{\pmod {n}}$, then:

• $(a_{1}+a_{2})\equiv (b_{1}+b_{2}){\pmod {n}}\,$
• $(a_{1}-a_{2})\equiv (b_{1}-b_{2}){\pmod {n}}\,$
• $(a_{1}a_{2})\equiv (b_{1}b_{2}){\pmod {n}}.\,$

Like any congruence relation, congruence modulo n is an equivalence relation, and the equivalence class of the integer a, denoted by ${\overline {a}}_{n}$, is the set $\left\{\ldots ,a-2n,a-n,a,a+n,a+2n,\ldots \right\}$. This set, consisting of the integers congruent to a modulo n, is called the congruence class or residue class of a modulo n. Another notation for this congruence class, which requires that in the context the modulus is known, is $\displaystyle [a]$.

The set of congruence classes modulo n is denoted as $\mathbb {Z} /n\mathbb {Z}$ (or, alternatively, $\mathbb {Z} /n$ or $\mathbb {Z} _{n}$) and defined by:

$\mathbb {Z} /n\mathbb {Z} =\left\{{\overline {a}}_{n}|a\in \mathbb {Z} \right\}.$

When n ≠ 0, $\mathbb {Z} /n\mathbb {Z}$ has n elements, and can be written as:

$\mathbb {Z} /n\mathbb {Z} =\left\{{\overline {0}}_{n},{\overline {1}}_{n},{\overline {2}}_{n},\ldots ,{\overline {n-1}}_{n}\right\}.$

We can define addition, subtraction, and multiplication on $\mathbb {Z} /n\mathbb {Z}$ by the following rules:

• ${\overline {a}}_{n}+{\overline {b}}_{n}={\overline {a+b}}_{n}$
• ${\overline {a}}_{n}-{\overline {b}}_{n}={\overline {a-b}}_{n}$
• ${\overline {a}}_{n}{\overline {b}}_{n}={\overline {ab}}_{n}.$

The verification that this is a proper definition uses the properties given before.

In this way, $\mathbb {Z} /n\mathbb {Z}$ becomes a commutative ring. For example, in the ring $\mathbb {Z} /24\mathbb {Z}$, we have

${\overline {12}}_{24}+{\overline {21}}_{24}={\overline {9}}_{24}$

as in the arithmetic for the 24-hour clock.

Properties of rings

We shall now discuss some basic theorems related to rings. We feel that a good way to learn ring theory is to try out proofs of simple theorems on ones own. Hence the reader is encouraged to work out proofs of theorems by him/herserlf before reading the proofs given here. Often we shall provide only a sketch of the proof and the reader is expected to fill in the gaps in that case.

Basic properties

Theorem 1.1: If R is a ring and $a,b,c\in R$; then

1. a+b=a+c implies b=c. (Cancellation Law)

2. -(-a)=a.

3. The zero element of R is unique.

4. The additive inverse of any element is unique.

Proof:

1. Clearly adding -a on both sides of a+b=a+c gives us the desired result.

2. It suffices to show that a+(-a)=0 which is obvious from the definition of -a.

3. If there exists two zero elements 0 and 0' in R then 0+0'=0' and 0+0'=0 by definition and so 0=0'.

4. If a' and a'' are two inverses of a then a'=a'+0=a'+a+a''=0+a''=a''.$\Box$

Theorem 1.2: If R is a ring, then for any $a,b,c\in R$;

1. a0=0a=0.

2. a(-b)=(-a)b=-(ab).

3. (-a)(-b)=ab.

4. a(b-c)=ab-ac.

If in addition, R has a unit element 1, then

5. (-1)a=-a.

6. (-1)(-1)=1.

Proof:

1. a0+0=a0=a(0+0)=a0+a0. By the cancellation law it now follows that a0=0. Similarly 0a=0.

2. It suffices to show that a(-b)=-(ab) or equivalently a(-b)+ab=0. Now a(-b)+ab=a(b-b)=a0=0 by 1. and so the result is proved.

3. (-a)(-b)=-(a(-b)) by 2. Again -(a(-b))=-(-(ab))=-(-ab) by 2. But by Theorem 1.1(2) -(-ab)=ab.

4. a(b-c)=a(b+(-c))=ab+a(-c)=ab-ac by 2.

5. (-1)a+a=(-1)a+1a=(-1+1)a=0a=0 by 1. and so (-1)a=-a.

6. Put a=-1 in 5. and apply Theorem 1.1(2).$\Box$

Some more results

It is strongly recommended that theorems in this section should be treated as exercises by the readers.

Theorem 1.3: Prove that a ring R is commutative if and only if $(a+b)^{2}=a^{2}+2ab+b^{2}$ holds for all $a,b\in R$.

Proof: Suppose R is commutative. Then clearly the result holds. (In fact the binomial theorem: $(a+b)^{n}=\textstyle \sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b^{k}$ holds in that case. Try to prove it using induction and the Pascal's identity:$\textstyle {\binom {n}{k}}+\textstyle {\binom {n}{k+1}}=\textstyle {\binom {n+1}{k+1}}$.) Conversely suppose that for each $a,b\in R$ the given relation is satisfied. Now, on applying the distributive laws to $(a+b)^{2}=(a+b)(a+b)$ we get $=a^{2}+ab+ba+b^{2}=(a+b)^{2}=a^{2}+ab+ab+b^{2}$ and by the cancellation laws we have ab=ba. Hence R is commutative.$\Box$

Theorem 1.4: If R is a system satisfying all the conditions of a ring with unit element with the possible exception of a+b=b+a, prove that the axiom a+b=b+a must hold in R and that thus R is a ring.

Proof: (a+b)(1+1)=a1+a1+b1+b1=a+a+b+b and (a+b)(1+1)=a1+b1+a1+b1=a+b+a+b by the left and right distributive laws reepectively. Equating the two identities and applying the cancellation laws gives us the result.$\Box$

Theorem 1.5: Let R be a ring such that $a^{2}=a$ for all $a\in R$. Prove that R is commutative.

Note: Such a ring is called a Boolean ring.

Proof: $(a+b)^{2}=a+b$ implies $a^{2}+ab+ba+b^{2}=a+b$. Since $a^{2}=a$ and $b^{2}=b$ so $ab=-ba$ by the cancellation law. Now as $a+a=(a+a)^{2}=a^{2}+2a^{2}+a^{2}=a+a+a+a$ so $a+a=0\ \forall a\in R$ and so each element in R is its own additive inverse. Hence $-ba=ba$ and so $ab=ba$.$\Box$

Theorem 1.6: If R is a ring with unity satisfying $(xy)^{2}=x^{2}y^{2}$ for all $x,y\in R$, prove that R is commutative.

Proof: By our hypothesis $[x(y+1)]^{2}=x^{2}(y+1)^{2}=x^{2}(y^{2}+2y+1)=x^{2}y^{2}+2x^{2}y+x^{2}$ and also by the distributive laws $[x(y+1)]^{2}=[xy+x]^{2}=(xy+x)(xy+x)=x^{2}y^{2}+xyx+x^{2}y+x^{2}$. So equating the two and applying the cancellation laws we have $xyx=x^{2}y$ which holds as an identity. Now substituting x+1 for x in the identity we have $(x+1)y(x+1)=(x+1)^{2}y$. This gives $(x+1)(yx+y)=(x+1)(xy+y)$ and on the application of the distributive laws we have $xyx+xy+yx+y=x^{2}y+xy+xy+y$. Cancellation law now gives $xy=yx$ as required.$\Box$

Theorem 1.7: Let R be a ring such that for $x\in R$, there exists a unique $a\in R$ such that xa=x. Show that ax=x. Hence deduce that if R has a unique right identity e, then e is the unity of R.

Proof: xa=x implies x(a+ax-x)=xa+xax-x²=x. Hence a+ax-x=a or ax=x. If R has a unique right identity e then xe=x implies ex=x and so e is the unity of R.$\Box$

Theorem 1.8: Let R be a ring with unity $1\in R$. Suppose for $x\neq 0\in R\exists$ a unique $y\in R$ such that xyx=x. Prove that xy=yx=1, i.e. x is invertible in R.

Proof: Suppose, if possible xa=0 for some $a\in R$. Now, x(y+a)x=(xy+xa)x=xyx+xax=xyx=x and by the uniqueness of y it follows that y+a=a i.e. a=0. So $xa=0\Rightarrow a=0$. Now x(yx-1)=xyx-x=x-x=0 and so yx-1=0. Hence yx=1. Similarly xy=1. So x is invertible.$\Box$

Theorem 1.9: Show that if 1-ab is invertible in a ring R with unity, then so is 1-ba.

Proof: Let x be the inverse of 1-ab, i.e. let x(1-ab)=(1-ab)x=1. Now (1-ba)(1+bxa)=1+bxa-ba-babxa=1-ba+b(1-ab)xa=1-ba+ba=1. Similarly (1+bxa)(1-ba)=1. So 1-ba is invertible with inverse 1+bxa.$\Box$

Theorem 1.10: If a,b are any two elements of a ring R and m and n are any two positive integers, then prove that

1. (m+n)a=ma+na.

2. m(a+b)=ma+mb.

3. m(na)=(mn)a.

4. (na)(mb)=(nm)(ab).

5. a^maⁿ=a^m+n.

6. (a^m)ⁿ=a^mn.

Proof We shall prove 4. and leave the rest as an exercise for the reader.

4. $(na)(mb)=(\underbrace {a+\cdots a} _{n\ times})(\underbrace {b+\cdots b} _{m\ times})=\underbrace {ab+\cdots ab} _{nm\ times}$ by repeated application of the distributive law. The RHS is just (nm)(ab).$\Box$

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

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

is a zero divisor because

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

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 $ab=0\Rightarrow a=0\ or\ b=0$ or in other words $a\neq 0,b\neq 0\Rightarrow ab\neq 0\forall a,b\in R$.

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

• Additive and multiplicative identity

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.

• Additive and multiplicative inverses

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 $\mathbb {Q} ,\mathbb {R}$ and $\mathbb {C}$. A division ring which is not a field is the field of quaternions, described as follows:

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

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

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

For example, since

$-1=ijk,$

right-multiplying both sides by k gives

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

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

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

For two elements $\alpha _{0}(1)+\alpha _{1}(i)+\alpha _{2}(j)+\alpha _{3}(k)$ and $\beta _{0}(1)+\beta _{1}(i)+\beta _{2}(j)+\beta _{3}(k)$, 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:

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

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 $ab=ac\Rightarrow b=c$ where $a,b,c\in R,a\neq 0$.

Proof: $\Rightarrow$: 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.

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

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 $a,b,c\in R$ and $a\neq 0$. Then as $a^{-1}$ 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.$\Box$

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

Theorem 1.13: Every finite integral domain is a field.

Proof: Let R be a finite integral domain and let $x\in R$ where $x\neq 0,1$. 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=x^j. Then xⁱ-x^j=0 and since i<j, so x^j-i is a legitimate member of R (in fact so is x^j-i-1). We have xⁱ(1-x^j-i)=xⁱ-x^j=0. As x is non zero and R is an integral domain so xⁱ is non zero. But then 1-x^j-i=0 or x^j-i=1. It follows that as x^j-i-1x=1. Hence x is a unit with inverse x^j-i-1.$\Box$

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

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

$\Leftarrow$ Suppose p is prime. It suffices to show that ${\frac {\mathbb {Z} }{p\mathbb {Z} }}$ is an integral domain. Let a,b be nonzero elements of ${\frac {\mathbb {Z} }{p\mathbb {Z} }}$ 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 ${\frac {\mathbb {Z} }{p\mathbb {Z} }}$ and so ${\frac {\mathbb {Z} }{p\mathbb {Z} }}$ is an integral domain.$\Box$

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

Subrings

Definition 1: Let (R,+,.) be a ring. A non empty subset S of R is called a subring of R if (S,+,.) is a ring.

For example the set $m\mathbb {Z}$ which stands for $\{0,\pm m,\pm 2m\cdots \}$ is a subring of the ring of integers, the set of Gaussian integers $\mathbb {Z} [i]$ is a subring of $\mathbb {C}$ and the set $\mathbb {Z} _{4}$ has the set $\{{\overline {0}},{\overline {2}}\}$ as a subring under addition and multiplication modulo 4.

Theorem 1.15: A non empty subset S of a ring R is a subring of R iff (i)$a-b\in S$ and (ii)$ab\in S\forall a,b\in S$.

Proof: The proof is an elementary consequence of a similar theorem about groups. Clearly necessity is clear. For sufficiency, note that 0=a-a is in S, -a=0-a is in S and a+b=a-(-b) is also in S. Other properties of a ring follow trivially.$\Box$

Theorem 1.16: The intersection of two subrings of a ring R is a subring of R.

Proof: Let S₁ and S₂ be two subrings and $a,b\in S_{1}\cap S_{2}$. Then $a-b,ab\in S_{1}\cap S_{2}$ as they are also in S₁ and S₂. (Note that the intersection is nonempty as it defnitely contains 0). But then the result holds by the previous theorem.$\Box$

Note that the corresponding result about unions may not be true. For example the union of $2\mathbb {Z}$ and $3\mathbb {Z}$ has 3 and 2 but not their difference 1 and so is not a subring of $\mathbb {Z}$.

Definition 2: The center of a ring R is defined as the set $Z(R)=\{a\in R:xa=ax\forall x\in R\}$.

Theorem 1.17: The center of a ring R is a subring of R.

Proof: Clearly if a and b are two elements in the center then for any x in R x(a-b)=xa-xb=ax-bx=(a-b)x and x(ab)=xab=axb=abx=(ab)x and so both a-b and ab are in the center. The result follows now from Theorem 1.15.$\Box$

Theorem 1.18: The center of a division ring is a field.

Proof: If R is a division ring, then its center contains the identity 1 as x1=1x=x for all x. Also if a is in the center and ab=ba=1 then for any x, xb=x1b=xabb=axbb and so x=axb. Now bx=baxb=1xb=xb and so b is also in the center. Hence each element's inverse is also in the center. Finally note that that the elements commute with each other as they do so with all other elements of R. Other properties of a field follow by virtue of R being a division ring.$\Box$.

Some more properties

These problems should be first tried as exercises by the reader.

Theorem 1.19: If a is a fixed element of a ring R, show that $I_{a}=\{x\in R:ax=0\}$ is a subring of R.

Proof: Clearly if x,y are two elements in R then a(x-y)=ax-ay=0-0=0 and a(xy)=axy=0y=0 and so I_a is a subring.$\Box$

Theorem 1.20:If A and B are two subrings of a ring R then their sum is defined as the set $A+B=\{a+b:a\in A,b\in B\}$. Show that the sum of two subrings need not be a subring.

Proof: Consider the ring M₂ of 2x2 matrices with entries belonging to the integers. (Check that this is indeed a ring.) It is easy to verify that the sets $S=\{{\bigl (}{\begin{smallmatrix}a&0\\b&0\end{smallmatrix}}{\bigr )}:a,b\in \mathbb {Z} \},T=\{{\bigl (}{\begin{smallmatrix}0&c\\0&0\end{smallmatrix}}{\bigr )}:c\in \mathbb {Z} \}$ are subrings of M₂. However their sum which contains the matrices ${\bigl (}{\begin{smallmatrix}1&1\\1&0\end{smallmatrix}}{\bigr )}$ and ${\bigl (}{\begin{smallmatrix}2&2\\2&0\end{smallmatrix}}{\bigr )}$ doesn't contain their product which is ${\bigl (}{\begin{smallmatrix}4&2\\2&2\end{smallmatrix}}{\bigr )}$. Hence sum of two subrings need not be a subring.$\Box$.

Theorem 1.21: An element $x\in R$ is called idempotent if x²=x. Let e be an idempotent in a ring R. Show that $eRe=\{eae:a\in R\}$ is a subring of R with unity e.

Proof: Clearly if x,y are in eRe then x=eae and y=ebe for some a,b in R. Then x-y=eae-ebe=e(a-b)e and so x-y is in eRe. Also xy=eaeebe=eaebe=e(aeb)e and so xy is in eRe. Hence eRe is a subring. Finally note that xe=eaee=eae=x and ex=eeae=eae=x and so e is the unity of eRe.$\Box$.

Exercises

1. Show that the normalizer N(a) of a element a of a ring R defined by $N(a)=\{x\in R:xa=ax\}$ is a subring of R.

2. A non empty subset S of a field (F,+,.) is called a subfield of F is (S,+,.) is a field. Show that a subset S of a field F, containing at least two elements is a subfield of F iff (i)$a-b\in S\forall a,b\in S$ and (ii)$ab^{-1}\in S\forall a\in S,b(\neq 0)\in S$.

Idempotent and Nilpotent elements

$x\in R$ is an Idempotent if $x^{2}=x$

$x\in R$ is nilpotent if $\exists n\in \mathbb {N}$ such that $x^{n}=0$