Page 55 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Corollary: Since integral domain has no zero divisors and field is necessarily an integral domain, Notes
therefore, field has no zero-divisor.
Theorem 16: If a, b are any two elements of a field F and a , there exists a unique element x such
that a . x = b.
–1
Proof: Let 1 be the unity of F and a , the inverse of a in F then
a . (a 1 ) b = (aa 1 ) . b 1 . b b

1
ax = b . a x a . (a b )
1
x a b (by left cancellation)
Thus x = a 1 b . F

Now, suppose there are two such elements x , x (say) then
1 2
a . x = b and .a x 2 b
1
Hence .a x 1 . a x 2

On applying left cancellation, we get
x = x
1 2
Hence the uniqueness is established.

Theorem 17: Every finite integral domain is a field.
or
A finite commutative ring with no zero divisor is a field.

Proof: Let D be an integral domain with a finite number of distinct elements a a 2 ,..., a In
,
.
n
1
order to prove that D is a field, we have to prove that there exists 1 D such that
1 . a a a D and for every a ( 0) D there exists an element a 1 D such that a 1 a 1.
Let a 0 and a D. Now the elements aa 1 aa 2 , ..., aa are the elements of D.
n
All of them are distinct because otherwise if aa i aa j , for i j then
aa = aa a(a – a ) = 0
i j i j
a – a = 0
i j
(because a 0 and D is without zero divisors)

a i a j contradicting i j.
Let one of these elements be a. Thus there exists an element, say 1 D such that

a . 1 a 1 . a (because multiplication is commutative)

Let y be any element of D then for some x D we should have
ax = y = xa
Therefore, 1y = 1 (ax) (because ax = y)
= (1a) x = ax (because 1a = a)

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