Group Ring Field-In Detail

Groups, rings, and fields are the fundamental elements of a branch of mathematics known as abstract algebra, or modern algebra. In abstract algebra, we are concerned with sets on whose elements we can operate algebraically; that is, we can combine two elements of the set, perhaps in several ways, to obtain a third element of the set. These operations are subject to specific rules, which define the nature of the set. By convention, the notation for the two principal classes of operations on set elements is usually the same as the notation for addition and multiplication on ordinary numbers.

Let's break down the concepts of groups, rings, and fields in the context of number theory, and then we'll discuss the notion of group rings and fields.

1. Groups: In mathematics, a group is a set equipped with an operation that combines two elements to produce a third, satisfying four fundamental properties:

Closure: The operation combines elements to produce another element in the set.

Associativity: The grouping of elements does not affect the result of the operation.

Identity Element: There exists an element that, when combined with any other element, leaves the latter unchanged.

Inverse Element: For every element, there exists an element that, when combined, results in the identity element.

Example: The integers under addition form a group. The identity element is 0, and each integer has an additive inverse.

2. Rings: A ring is a set equipped with two operations (usually addition and multiplication) that satisfies several properties:

The set forms an abelian group under addition.

Multiplication is distributive over addition.

There is an associative multiplication.

Example: The set of integers with the usual addition and multiplication forms a ring.

3. Fields: A field is a special type of ring where multiplication is commutative, and every non-zero element has a multiplicative inverse.

Example: The set of rational numbers, real numbers, and complex numbers are all examples of fields.

$(R; +, ·)$ and $(Q; +, ·)$ serve as examples of fields, $(Z; +, ·)$ is an example of a ring which is not a field.

Finite field $\mathbb{F}_2​$ with two elements, 0 and 1. This field is commonly used in binary systems. Elements in $\mathbb{F}_2$​ are operated upon using modular arithmetic with respect to 2 (addition and multiplication are performed modulo 2).

Lets understand these concepts in detail

Binary operations(bop)

Let $S$ be a non-empty set. A map 

$$ ⋆: S × S → S, (a, b) → a ⋆ b$$ is called a binary operation on $S$. So ⋆ takes 2 inputs $a, b$ from $S$ and produces a single output $a ⋆ b ∈ S$. In this situation we may say that ‘$S$ is closed under ⋆’. 

Aside: A unary operation on a non-empty set $S$ is a map from $S$ to $S$. Examples are $a → (−a) \quad on \quad\mathbb{Z}$, 

$a → 2a \quad  on \quad \mathbb{C}$. 

Let ⋆ be a binary operation on a set $S$. We say  ⋆ is commutative if, for all $a, b ∈ S, a ⋆ b = b ⋆ a$. 

⋆ is associative if, for all $a, b, c ∈ S, a ⋆ (b ⋆ c) = (a ⋆ b) ⋆ c$ (note that (bop) ensures that each side of this equation makes sense). If ⋆ is associative we can unambiguously write $a ⋆ b ⋆ c$ to denote either of the iterated products. Very convenient. 

Provable fact: Let ⋆ be an associative binary operation on a set $S$ and let $x_1, . . . , x_n ∈ $. Then $x_1 ⋆ x_2 ⋆ . . . ⋆ x_n$ can be unambiguously defined.

Examples of binary operations. 

(1) Addition, +, is a commutative and associative binary operation on each of the following: 

$\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}, M_{m,n}(R) (m, n\ge 1)$, real polynomials.

(2) Multiplication, $\cdot$ , is an associative and commutative binary operation on each of the following:

 $\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$,real polynomials.

Matrix multiplication is an associative binary operation on $M_n(R)$; for $n > 2$ this is NOT commutative.

The definition of a group. 

Let $G$ be a non-empty set and let ⋆ be a binary operation on $G$ $$ ⋆: G × G → G, (a, b) → a ⋆ b$$. Then $(G; ⋆)$ is a group if the following axioms are satisfied: 

(G1) associativity: $a ⋆ (b ⋆ c) = (a ⋆ b) ⋆ c$ for all $a, b, c ∈ G$

(G2) identity element: there exists $e ∈ G$ such that $a ⋆ e = e ⋆ a = a$ for all $a ∈ G$.

(G3) inverses: for any $a ∈ G$ there exists $a −1 ∈ G$ such that$ a ⋆ a^{−1} = a ^{−1} ⋆ a = e$. If in addition the following holds

(G4) commutativity: $a ⋆ b = b ⋆ a$ for all $a, b ∈ G$ then $(G; ⋆) $is called an abelian group, or simply a commutative group. If the set $G$ is finite, we define the order of $G$ to be the number of elements in $G$, and denote it $|G|$. Otherwise we say that $G$ has infinite order. 

Fact: if $(G; ⋆)$ is a group then the identity $e$ is unique and the inverse of any a in $G$ is uniquely determined by $a$.

Examples of groups 

(1) $(Z; +), (Q; +), (R; +), (C; +)$ are abelian groups. 

(2) Let $V$ be a real vector space.  $(V ; +)$ is an abelian group.

(3) $(Q \setminus \{0\}, ·), (R \setminus \{0\}, ·), (C \setminus \{0\}, ·)$ are abelian groups. 

(4) The invertible $n × n$ complex matrices form a group under matrix multiplication. This is an important group, denoted $GL(n, C)$ or $GL_n(C)$. For $n > $1 this group is non-abelian.

When the group operation is addition, the identity element is 0; the inverse element of $a$ is $–a$ ; and subtraction is defined with the following rule:

$a-b=a+(-b)$

Cyclic Group

We define exponentiation within a group as a repeated application of the group operator, so that .$a^3=a·a·a$ Furthermore, we define $a^0$ as the identity element, and , $a^{-n}=(a^{-1})^n$,where $a^{-1}$ is the inverse element of within the group.A group is cyclic if every element of $G$ is a power of $a^k$ ( $k$ is an integer) of a fixed element . The element $a$ is said to generate the group or to be a generator of $G$. A cyclic group is always abelian and may be finite or infinite.

The definition of a ring. 

A structure $(R, +, ·)$ is a ring if $R$ is a non-empty set and + and · are binary operations: 

$$+: R × R → R, (a, b) → a + b$$ 

$$· : R × R → R, (a, b) → a · b $$

such that 

Addition: $(R, +)$ is an abelian group, that is, 

(A1) associativity: for all $a, b, c ∈ R$ we have $a + (b + c) = (a + b) + c$ 

(A2) zero element(identity for addition): there exists $0 ∈ R$ such that for all $a ∈ R$ we have $a + 0 = 0 + a = a$

(A3) inverses: for any $a ∈ R$ there exists $−a ∈ R$ such that $a + (−a) = (−a) + a = 0$ 

(A4) commutativity: for all $a, b ∈ R$ we have $a + b = b + a$

Multiplication: 

(M1) associativity: for all $a, b, c ∈ R$ we have $a · (b · c) = (a · b) · c$ 

Addition and multiplication together ( distributive)

(D) for all $a, b, c ∈ R, a · (b + c) = a · b + a · c$ and $(a + b) · c = a · b + b · c.$

We sometimes say ‘$R$ is a ring’, taken it as given that the ring operations are denoted + and ·. As in ordinary arithmetic we shall frequently suppress · and write $ab$ instead of $a · b$. 

We do NOT demand that multiplication in a ring be commutative.

Special types of rings: definitions. 

Assume $(R; +, ·)$ is a ring. We say $R$ is a commutative ring if its multiplication · is commutative, that is, 

(M4) Commutativity: $a · b = b · a$ for all $a, b ∈ R$. We say $R$ is a ring with 1 (or ring with identity) if there exists an identity for multiplication, that is, 

(M2) identity element: there exists $1 ∈ R$ such that for all $a ∈ R$ we have $a · 1 = 1 · a = a.$

Examples of rings. 

Number systems 

(1) All of $Z, Q, R$ and $C$ are commutative rings with identity (with the number 1 as the identity). (2) $N$ is NOT a ring for the usual addition and multiplication. These are binary operations and we do have a zero element, namely 0, so axiom (A2) holds. However (A3) (existence of additive inverses) fails: there is no $n ∈ N$ for which $1 + n = 0$, for example. . 

(3) Consider the set of even integers, denoted $2Z$, with the usual addition and multiplication. This is a commutative ring without an identity. To verify that (M2) fails it is not sufficient just to say that the integer 1 does not belong to $2Z$. Instead we argue as follows. Suppose for contradiction that there were an element $e ∈ 2Z$ such that $n·e = n$ for all $n ∈ 2Z$. In particular $2e = 2$, from which we deduce that $e$ would have to be 1. Since $1 \notin 2Z$ we have a contradiction.

Matrix rings Under the usual matrix addition and multiplication $M_n(R)$ and $M_n(C)$, are rings with 1, but are not commutative (unless n = 1). If we restrict to invertible matrices we no longer have a ring, because there is then no zero for addition. 

Polynomials Polynomials, with real coefficients, form a commutative ring with identity under the usual addition and multiplication; we denote this by $R[x]$. 

Modular arithmetic Binary arithmetic on $\{0, 1\}$  gives us a 2-element commutative ring with identity. More generally we get a commutative ring with identity if we consider addition and multiplication mod $n$ on $\{0, 1, . . . , n − 1\}$. 

Calculational rules for rings. Assume that $(R; +, ·)$ is a commutative ring. Let $a, b, c ∈ R$. 

(i) If $a + b = a + c$ then $b = c$. 

(ii) If $a + a = a$ then $a = 0$. 

(iii)$ −(−a) = a$. 

(iv) $0a = 0$. 

(v) $−(ab) = (−a)b = a(−b)$. Assume in addition that $R$ has an identity 1 Then 

(vi) $(−1)a = −a$. 

(vii) If $a ∈ R$ has a multiplicative identity $a ^{−1}$ then $ab = 0$ implies $b = 0$.

Subrings and the Subring Test. Let $(R; +, ·)$ be a ring and let $S$ be a non-empty subset of $R$. Then $(S; +, ·)$ is a subring of $R$ if it is a ring with respect to the operations it inherits from $R$. 

The Subring Test Let $(R; +, ·)$ be a ring and let $S ⊆ R$. Then $(S; +, ·)$ is a subring of $R$ if (and only if) $S$ is non-empty and the following hold: 

(SR1) $a + b ∈ S$ for any $a, b ∈ S$; 

(SR2) $a − b ∈ S$ for any $a, b ∈ S$; 

(SR3) $ab ∈ S$ for any$ a, b ∈ S$. 

Here (SR1) and (SR3) are just the (bop) conditions we require for $S$. (A1), (M1) and (D) are inherited from $R$. The only (∃) axioms are (A2) and (A3). Since $S\ne ∅$ we can pick some $c ∈ S$. Then, by (SR2), $0 = c − c ∈ S$. By (SR2) again, $−a = 0 − a ∈ S$. 

Examples 

(1) $Z$ and $Q$ are subrings of $R$; 

(2) $R$, regarded as numbers of the form $a + 0i$ for $a ∈ R$, is a subring of $C$. 

(3) In the polynomial ring $R[x]$, the polynomials of even degree form a subring but the polynomials of odd degree do NOT form a subring because $x · x = x^2$ is not of odd degree. 

(4) $nZ := \{nk | k ∈ Z\}$ is a subring of $Z$ for any $n ∈ N$.

 

Fields and integral domains 

A commutative ring with identity, $(R, +, ·)$ satisfies all of the axioms (A1)–(A4), (that is, $(R, +)$ is an abelian group); (M1), (M2), (M4); (D) (distributivity). 

What’s ‘missing’ here is (M3) inverses: for all $a ∈ R$ with $a \ne 0$ there exists $1/a ∈ R$ (alternatively written $a −1$ ) such that $a · 1/a = 1/a · a = 1$. 

Definition of a field: A structure $(R, +, ·)$, where + and · are binary operations on $R$ is a field if (A1)–(A4), (M1)–(M4), and (D) hold, and $0 \ne 1$. This can be expressed in a more modular way as follows $(R, +, ·)$ is a field if 

(A) $(R, +)$ is an abelian group; 

(M) $R \setminus \{0\}, ·)$ is an abelian group; 

(D) the distributive laws hold.

Examples of fields: The following are fields: $Q, R, C$. 

The following commutative rings with identity fail to be fields: $Z, K[x]$; here $K$ denotes $R$, or $C$, or any other field, and $K[x]$ the polynomials with coefficients in $K$. 

Definitions: In a commutative ring we call an element $a \ne 0$ a zero divisor if there exists $b \ne 0$ such that $a · b = 0$. A commutative ring with identity in which $0 \ne 1$ is an integral domain (ID) if it has no zero divisors. 

Observation (cancellation property): in an integral domain $ab = ac$ and $a \ne 0$ implies $b = c$. 

Examples of integral domains 

(1) We claim that any field is an integral domain. To prove this, assume that $(R, +, ·)$ is a field and let $a, b ∈ R$ be such that $a · b = 0$. If $a \ne  0$ then $a^{ −1}$ exists, and we have $$0 = a^{ −1} · 0 = a^{ −1}· (a · b) = (a^{ −1} · a) · b = 1 · b = b$$. and likewise with the roles of $a$ and $b$ reversed.

(2) $Z$ and $K[x]$ are integral domains which fail to be fields. 

(3) $R^2$ , with coordinatewise addition and multiplication  is a commutative ring with identity which fails to be an integral domain (and so is not a field): (0, 1) · (1, 0) = (0, 0).

Theorem. A finite integral domain is a field. 

Proof . Let $R$ be a finite integral domain and list its distinct elements as $a_1, . . . , a_n$. Let $b ∈ R, b \ne 0$. We have to show there exists $c ∈ R$ such that $bc = 1$. To do this we invoke the cancellation property of an ID: $ba_i = ba_j$ implies $a_i = a_j$ . Therefore $\{ba_1, . . . , ba_n\}$ contains $n$ elements and is all of $R$. In particular, there exists $j$ such that $ba_j = 1$.

Units. Let $R$ be a commutative ring with 1. Then $0 \ne u ∈ R$ is a unit if there exists $v ∈ R$ such that $uv = vu = 1$; we write as usual $v = u ^{−1}$ . Thus the units are those elements (necessarily non-zero) which have multiplicative inverses.

Examples 

(1) In a field, every non-zero element is a unit. 

(2) In $Z$, the units are ±1. 

(3) In the ring of real or complex polynomials, or more generally the ring $K[x]$ of polynomials over a field $K$, the units are the non-zero constant polynomials. 

Summing up what we have about $Z$ so far.:

Theorem (the integers as a ring). 

(i) $Z$ is a commutative ring with 1. 

(ii) $n ∈ Z$ has a multiplicative inverse in $Z$ (that is, $n$ is a unit) if and only if $n = ±1$. 

(iii) $Z$ fails to be a field, but is an integral domain: (ID) ab = 0 implies a = 0 or b = 0.

Difference Between Field and Integral Domain

Both integral domains and fields are algebraic structures, but they differ in their level of "closeness" with respect to multiplication. Let's explore the definitions and key differences between integral domains and fields:

Integral Domain:

An integral domain is a commutative ring with unity (a ring with addition and multiplication) in which the product of any two non-zero elements is non-zero. In other words, it has no zero divisors. More formally, a set $D$ with operations of addition and multiplication is an integral domain if:

It is a commutative ring with unity.

For any non-zero elements $a$ and $b$ in $D$, the product $ab$ is non-zero.

Example: The set of integers ($Z$) is an integral domain because the product of any two non-zero integers is non-zero.

Field:

A field is a special type of integral domain in which every non-zero element has a multiplicative inverse. In other words, every non-zero element has a reciprocal such that when multiplied, they give the multiplicative identity. Formally, a set $F$ with operations of addition and multiplication is a field if:

It is a commutative ring with unity.

Every non-zero element in $F$ has a multiplicative inverse.

Example: The set of real numbers ($R$), the set of complex numbers ($C$), and finite fields like $Z_p​$ (where $p$ is a prime number) are examples of fields.

Key Differences:

Existence of Multiplicative Inverses:

In an integral domain, there is no requirement for every non-zero element to have a multiplicative inverse.

In a field, every non-zero element must have a multiplicative inverse.

Closure under Multiplication:

In both integral domains and fields, multiplication is closed, associative, and commutative.

The crucial difference lies in whether every non-zero element has a reciprocal.

Zero Divisors:

Integral domains don't have zero divisors, meaning that if the product of two elements is zero, then at least one of the elements must be zero.

Fields, being integral domains, also don't have zero divisors.

In summary, a field is a special case of an integral domain where every non-zero element has a multiplicative inverse. Fields have a higher degree of symmetry under multiplication compared to integral domains. Both structures are essential in algebra and have applications in various mathematical and scientific fields.