Continuing with previous notes of Abstract Algebra, here we come to the concept of ** Rings**. This time we use the textbook of

**.**

*Tom Judson’s Abstract Algebra: Theory and Applications*## Rings

A ring is a non-empty set R that has two closed binary operations, addition and multiplication, satisfying the following conditions:

**Abelian group**: with addition as binary operation, R is an abelian group, with 0 as the identity.**Associative**: with multiplication as binary operation, \forall a\ b\ c \in R, (ab)c=a(bc)**Distributive**: combining addition and mutiplication, \forall a\ b\ c \in R:- a(b+c) = ab+ac
- (a+b)c = ac+bc.

In mathematics, rings are algebraic structures that generalize fields: multiplication need not be commutative and multiplicative inverses need not exist.

From Wikipedia, the free encyclopedia

A unit is an element that has a multiplicative inverse. A nonzero element a in a commutative ring R is called a zero divisor if there is a nonzero element b in R such that ab = 0.

Rings with more properties:

**A ring with identity**: \exists 1 \in R, \text{ such that } 1\ne 0 \text{ and } \forall a \in R, 1a=a1=a.**Commutative ring**: mutiplication satisfies commutative law: \forall a\ b \in R , ab=ba.**Integeral domain**: a commutative ring with identity, also with the condition \forall a \ b \in R , ab=0 \implies a = 0 \lor b = 0.**Division ring**: a ring with identity, and every non-zero element is a unit, means \forall a \in R, a \ne 0, \exists a^{-1} \in R, \text{ such that } a^{-1}a=aa^{-1} = 1.**Field**: A commutative division ring.

**Proposition:** Let R be a ring with a,b\in R. Then:

- a0=0a=0;
- a(-b)=(-a)b=-ab;
- (-a)(-b)=ab.

*Proof*:

(1) Note that 0 is the identity under addition, so that 0+0=0, then

a0 = a(0+0) = a0+a0 \implies a0 = 0Same is for 0a.

(2) Note that -b is the inverse under addition, so that

a(b + (-b)) = ab + a(-b) = a0 = 0 \implies ab \text{ is the inverse of } a(-b) .

Also, we have

(a+(-a))b = ab + (-a)b = 0b = 0 \implies ab \text{ is the inverse of } (-a)b .

Finally we have -(ab)=(-a)b=a(-b).

(3) Based on (2), (-a)(-b) = (-(-a))b=ab. *QED*

A **subring** S of a ring R is:

- S is a subset of R
- S is also a ring with the operations inherited from R

Abstract Algebra

Notice that even though the original ring may have an identity, we do not require that its subring have an identity.

**Proposition:** Let R be a ring and S be a subset of R. Then S is a subring of R *if and only if**the following conditions are satisfied*:

- S \ne \emptyset
- \forall r\ s \in S, rs \in S
- \forall r\ s \in S, r-s \in S.

*Proof*:

(1) \implies direction: this is quite simple, since subring’s definition requires closed addition and mutiplication.

(2) \Longleftarrow direction:

\text{with 1} \implies \text{ at least } r \in S ,

\text{ with 3 } \implies r - r \in S \implies 0 \in S,

0\in S, \text{ with 3 } \implies \forall s \in S, -s \in S \implies \text{ inverse exists and closure },

\forall r\ s \in S, -s \in S, r - (-s) = r + s \in S \implies \text{ addition closure }, remember the binary operations inherited guarantee associative and communicate.

\text{with 2} \implies \text{ mutiplication closure }.

Distribution law automatically closures with all the above. ** QED**.

## Integral Domains and Fields

If R is a commutative ring and r is a non-zero element in R, then \exists s\ne 0 \in R, rs = 0 \implies r \text{ is a zero divisor}.

R \text{ is a commutative ring with identity, and has no zero divisors} \implies R \text{ is an integral domain}.R \text{ is a ring with identity,} r \in R, \text{ r has mutiplication inverse } \implies r \text{ is a unit} . \forall r \ne 0 \in R, r \text{ is a unit} \implies R \text{ is a division ring}.

A commutative division ring is called **field**.

**Gaussian Integers:** \mathbb{Z}[i] = \{m+ni:m, n \in \mathbb{Z}\}. It is a subring of complex number. It is not a field, but an integral domain.

** Proof**:

(1) **integral domain:** identity and commutative is simple to verify, let’s focus on no zero divisors.

(2) **not a field:** easy to show 5 does not have multiplication inverse in \mathbb{Z}[i].

*QED.*

**Cancellation Law:** Let D be* *a commutative ring with identity. D \text{ is an integral domain } \iff \forall a \ne 0 \in D, ab=ac \implies b=c.

*Proof:*

(1) \implies direction:

a\ne 0, ab=ac \implies ab+(-ac) = a(b-c)=0 \land D \text{ is integral domain } \implies b-c=0 \implies b = c(2) \Longleftarrow direction:

ab = 0 \implies a = 0 \lor (ab = a0 \land a \ne 0) \implies a = 0 \lor b=0.

*QED.*

**Theorem:** Every finite integral domain is a field.

*Proof:*

Let D be a finite integral domain, and D^{*} be the set of nonzero element in D. For each element in D^{*}, we can define a map \forall a \in D^{*}, \lambda_{a} : D^{*} \mapsto D^{*} \text{ by } \lambda_{a}(d)= ad.

The above map is well-defined: a \ne 0, d \ne 0 \implies ad \ne 0. And this map is bijective:

- injective: ad_1 = ad_2 \land a \ne 0 \implies d_1 = d_2 \text{ (using the cancellation law) } ;
- surjective: \text{obvious????}

Thus \exists d \in D^{*} ad = 1, a has a left inverse. Based on commutative, d must also be a right inverse for a.

So it is a field.

*QED.*

**Characteristic of a ring:** For any nonnegative integer n and any element r \in \text{ a ring }R, we write r+r+\dots + r(n \text{ times as }) nr. Then

\text{characteristic of a ring }R = \arg \min_{n} \forall r \in R, nr = 0 \land n > 0 .

If the above condition cannot be satisfied, then characteristic of the ring is 0.

**Lemma: **Let R be a ring with identity. Then \text{order of } 1 = n\implies char(R) = n .

*Proof:*

1 has order n, so that n is the least positive integer n1=0. Then let’s check other elements. \forall r \in R, nr = n(1r) = (n1)r = 0 r = 0.

*QED.*

**Theorem:** The characteristic of an integral domain is either prime or zero.

*Proof:*

Let D be an integral domain and suppose that the characteristic of D is n with n \ne 0. If n is not prime, then n=ab, 1<a<n,1<b<n.

n1=ab1=(a1)(b1) = 0 \land \text{ integral domain } \implies a1 = 0 \lor b1 = 0. So that the characteristic of D must be less than n. Proof by contradiction.

*QED.*

## Ring Homomorphisms and Ideals

A homomorphism between rings preserves the operations of addition and multiplication in the ring.

If R \text{ and } S \text{ are rings, then a ring homomorphism is a map }\phi: R \mapsto S satisfying:

\forall a\ b \in R, \phi(a+b)=\phi(a)+\phi(b), \phi(ab)=\phi(a)\phi(b)If \phi is a bijective map, then it is called is isomorphism.

Kernel is a very important concept in the theory of ring. For any ring homomorphism \phi : R \mapsto S, we define the kernel of a ring homomorphism to be the set:

ker\ \phi = \{ r \in R \mid \phi (r) = 0 \} .

**Example:** For any integer n we can define a ring homomorphism \phi: \mathbb{Z} \mapsto \mathbb{Z}_n by a \mapsto a \pmod{n} . We can verify that this is a ring homomorphism and its kernel is ker\ \phi = n\mathbb{Z} .

**Evaluation homomorphism:** C[a,b] is the ring of continuous real-valued functions on an interval [a,b]. Given a fixed value \alpha \in [a,b], define a ring homomorphism \phi_{\alpha}: C[a,b] \mapsto \mathbb{R} \text{ by } \phi_\alpha(f) = f(\alpha).

**Proposition of ring** **homomorphism:** Let \phi : R \mapsto S be a ring homomorphism:

- If R is a commutative ring, then \phi (R) is also a commutative ring
- \phi (0) = 0
- Let 1_R \text{ and } 1_S be the identities for R \text{ and } S, then \phi \text{ is surjective } \implies \phi (1_R) = 1_S.
- R \text{ is a field } \land \phi(R) \ne \{0\} \implies \phi(R) \text{ is a field.}

In group theory we found that normal subgroups play a special role. These subgroups have nice characteristics that make them more interesting to study than arbitrary subgroups. In ring theory the objects corresponding to normal subgroups are a special class of subrings called ideals.

**Ideal:** an ideal in a ring R is a subring I of R such that a \in I \land r \in R \implies ar \in I \land ra \in I. Or we can say: \forall r \in R, rI = I \land Ir = I.

Let R be a ring with identity and suppose that I is an ideal of R and 1\in I. By definition, we have \forall r \in R, 1r \in I, so that I=R.

**Example:** Let a be any element in a commutative ring R, then the set defined as \langle a \rangle = \{ar \mid r\in R\} . Then this set is an ideal. This is easy to verify.

If R is a commutative ring with identity, then an ideal that has the form \langle a \rangle = \{ar \mid r \in R\} is called a principal ideal.

**Theorem :** Every ideal in the ring of integers \mathbb{Z} is a principal ideal.

*Proof:*

If the ideal only contains 0, I = \{0\}, it is a trivial ideal has the form \langle 0 \rangle. Then let’s consider that ideals that contain elements other than 0. Such ideal Imust contain at least one positive integer, and let n be the least positive integer in it. Let a \in I, a = nq + r, r = a \mod n, 0 \le r < n. Then r = a - nq , we know n \in I, a \in I \text{ so that } r \in r , together with n is the least positive element in I, then r=0. So we know \forall a \in I, a = nq , which means I = \langle n \rangle.

*QED.*

**Proposition:** The kernel of any ring homomorphism \phi : R \mapsto S is an ideal in R.

*Proof:*

From group theory, ker\ \phi is additive subgroup of R. Given \forall r \in R, a \in ker\ \phi:

\phi (ar) = \phi(a)\phi(r) = 0\phi(r) = 0, \phi(ra) = 0*QED.*

**Theorem:** Let I be an ideal of R. Then the factor group R/I is a ring with mutiplication defined as

## Maximal and Prime Ideals

**maximal ideal:** A proper ideal M of a ring R, and \forall S \subset R \land S \ne R \land S\text{ is ideal }, M \not \subset S.

**Theorem:** Let R be a commutative ring with identity and M an ideal in R. Then M \text{ is maximal ideal } \iff R/M \text{ is a field} .

**Theorem:** Let R be a commutative ring with identity 1 , where 1 \ne 0 . Then P \text{ is a prime ideal in } R \iff R/P \text{ is an integral domain} .