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show that if A is an ordered integral domain, then max A and min A do...

show that if A is an ordered integral domain, then max A and min A do not exist. What does this imply about finite integral domains?

thanks for the help.

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Answer #1

If possible let max(A) exists. Let a = max(A).

Since A is an integral domain, so it is a ring , also it is a group and therefore it satisfies the closure property.

So, by closure property, a + a belongs to A and a - a belongs to A. That is 2a belongs to A and 0 belongs to A. Then a fails to be the maximum max(A).

Similarly if possible let b= min(B).

By closure property b - b = 0 belongs to A, and b + b = 2b belongs to A.

Again b fails to be the minimum of A.

So , max(A) and min(A) doesn't exists.

Second part :

Let the Integral domain is finite. If it is ordered integral domain then there will be no maximum and no minimum and therefore it can not be finite. So, the finite integral domains are not ordered. Otherwise it will be infinite.

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