Page 54 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes (F ) There exists an identity element 1 for multiplication F such that
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a 1 1 a a a F.
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(F ) For all a F, a 0, there exists an element a (multiplicative inverse) in F such that
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a a 1 a 1 a 1 .
F . Distributive laws of multiplication over addition for all , ,a b c F ,
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a (b c ) a b a c
and (b c a b a c a
)
The above properties can be summarised as:
(1) (F, +) is an abelian group.
(2) (F, .) is a semi-abelian group and (F – {0}, .) is an abelian group.

(3) Multiplication is distributive over addition.

Examples:
(i) The set of real numbers is a field under usual addition and multiplication compositions.
(ii) The set of rational numbers is a field under usual addition and multiplication operations.
(iii) The set of integers is not a field.

Some Theorems

Theorem 14: The multiplicative inverse of a non-zero element of a field is unique.
Proof: Let there be two multiplicative inverse a and a for a non-zero element a F.
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Let (1) be the unity of the field F.
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aa = 1 and a . a = 1 so that a . a = a . a .
Since F – {0} is a multiplicative group, applying left cancellation, we get a = a .
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Theorem 15: A field is necessarily an integral domain.
Proof: Since a field is a commutative ring with unity, therefore, in order to show that every field
is an integral domain we only need proving that a field is without zero divisors.

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Let F be any field let a, b F with a 0 such that ab = 0. Let 1 be the unity of F. Since a 0, a exists
in F and therefore,
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ab = 0 a (ab) = a 0
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(a a) b = 0 (because a a =1)
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1 b = 0
b = 0 (because 1. b = b)
Similarly if b 0 then it can be shown that
ab = 0 a = 0

Thus ab = 0 a = 0 or b = 0.
Hence, a field is necessarily an integral domain.

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