Define (i) group (ii) Ring (iii) Field
Question
Define (i) group (ii) Ring (iii) Field
Solution
(i) Group: A group is a set, G, together with an operation • (called the group law of G) that combines any two elements a and b to form another element, denoted a • b or ab. To qualify as a group, the set and operation, (G, •), must satisfy four requirements known as the group axioms:
- Closure: For all a, b in G, the result of the operation, or "product", a • b, is also in G.
- Associativity: For all a, b and c in G, (a • b) • c = a • (b • c).
- Identity element: There is an element e in G such that, for every element a in G, the equations e • a and a • e return a.
- Inverse element: For each a in G, there exists an element b in G, commonly denoted a^−1 (or −a, if the operation is denoted "+"), such that a • b = e and b • a = e.
(ii) Ring: A ring is a set R equipped with two binary operations + and · satisfying the following three sets of axioms, called the ring axioms:
- R is an abelian group under addition, meaning that: (a) (a + b) + c = a + (b + c) for all a, b, c in R (that is, + is associative). (b) a + b = b + a for all a, b in R (that is, + is commutative). (c) There is an element 0 in R such that a + 0 = a for all a in R (that is, 0 is the additive identity). (d) For each a in R there exists −a in R such that a + (−a) = 0 (that is, −a is the additive inverse of a).
- R is a monoid under multiplication, meaning that: (a) (a · b) · c = a · (b · c) for all a, b, c in R (that is, · is associative). (b) There is an element 1 in R such that a · 1 = a and 1 · a = a for all a in R (that is, 1 is the multiplicative identity).
- Multiplication is distributive with respect to addition, meaning that: (a) a · (b + c) = (a · b) + (a · c) and (b + c) · a = (b · a) + (c · a) for all a, b, c in R.
(iii) Field: A field is a set F equipped with two operations called addition and multiplication. A field is essentially a ring in which every non-zero element has a multiplicative inverse. So, it satisfies all the axioms of a ring and additionally:
- The multiplication operation is commutative, i.e., for all a and b in F, a · b = b · a.
- There is an element 1 ≠ 0 in F, such that for every element a in F, 1 · a = a · 1 = a.
- For every non-zero element a in F, there exists an element b in F such that a · b = b · a = 1.
- The distributive law: For all a, b and c in F, a · (b + c) = (a · b) + (a · c).
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