prove that every finite integral domain is a field .is it true for infinite integral domain ? justify the answer
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hey friend here your answer
If a, b are two ring elements with a, b ≠ 0 but ab = 0 then a and b are called zero-divisors.
Example
In the ring Z6 we have 2.3 = 0 and so 2 and 3 are zero-divisors.
More generally, if n is not prime then Zncontains zero-divisors.
Proof
The only thing we need to show is that a typical element a ≠ 0 has a multiplicative inverse.
Consider a, a2, a3, ... Since there are only finitely many elements we must have am = an for some m < n(say).
Then 0 = am - an = am(1 - an-m). Since there are no zero-divisors we must have am ≠ 0 and hence 1 - an-m = 0 and so 1 = a(an-m-1) and we have found a multiplicative inverse for a.
I hope it's help you
mark brainiest
If a, b are two ring elements with a, b ≠ 0 but ab = 0 then a and b are called zero-divisors.
Example
In the ring Z6 we have 2.3 = 0 and so 2 and 3 are zero-divisors.
More generally, if n is not prime then Zncontains zero-divisors.
Proof
The only thing we need to show is that a typical element a ≠ 0 has a multiplicative inverse.
Consider a, a2, a3, ... Since there are only finitely many elements we must have am = an for some m < n(say).
Then 0 = am - an = am(1 - an-m). Since there are no zero-divisors we must have am ≠ 0 and hence 1 - an-m = 0 and so 1 = a(an-m-1) and we have found a multiplicative inverse for a.
I hope it's help you
mark brainiest
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