The multiplicative inverse of a non-zero element of a field is unique.
Let there be two multiplicative inverse and for a non-zero element . Let be the unity of the field .
and so that . Since is a multiplicative group, applying left cancellation, we get .
A field is necessarily an integral domain.
Since a field is a commutative ring with unity, therefore, in order to show that every field is an integral domain we only need to prove that s field is without zero divisors.
Let be any field and let with such that . Let be the unity of . Since , exists in , therefore
Similarly if then it can be shown that .
Thus . Hence, a field is necessarily an integral domain.
Since the integral domain has no zero divisor and the field is necessarily an integral domain, therefore the field has no zero-divisor.
If are any two elements of a field and , there exists a unique element such that .
Let be the unity of and , the inverse of in then
Now, suppose there are two such elements (say), then and hence . On applying left cancellation, we get .
Hence the uniqueness is established.
Every finite integral domain is a field, or a finite commutative ring with no zero divisor is a field.