characterization of prime ideals
We start with a general ring .
Let such that . Let and be the (left, right or two-sided) ideals generated by and , respectively. Then each element of the product of ideals can be expanded to a finite sum of products each of which contains or is a factor of the form for a suitable . Since is an ideal and , it follows that . Assuming statement 1, we have , or . But , so we have or and hence or .
Let be (left, right or two-sided) ideals, such that the product of ideals . Now or (depending on what type of ideal we consider), so . If , nothing remains to be shown. Otherwise, let , then for all . Since we have by statement 2 that for all , hence .
There are some additional properties if our ring is commutative.
Let a commutative ring and an ideal. Then the following statements are equivalent:
Given ideals of such that the product of ideals , then or .
Given such that , then or .
Let be arbitrary nonzero elements. Let and be representatives of and , respectively, then and . Since is commutative, each element of the product of ideals can be written as a product involving the factor . Since is an ideal, we would have if which by statement 1 would imply or in contradiction with and . Hence, and thus .
Let such that . If both were not elements of , then by statement 3 would not be an element of . Therefore at least one of is an element of .
Let be ideals of such that . If , nothing remains to be shown. Otherwise, let . Then for all the product , hence . It follows by statement 4 that , and therefore .
The condition 4 that the set is a multiplicative semigroup. Now is trivially the greatest ideal which does not intersect .
We presume that is maximal of the ideals of which do not intersect a semigroup and that . Assume the contrary of the assertion, i.e. that and . Therefore, is a proper subset of both and . Thus the maximality of implies that
So we can choose the elements and of such that
where , and . Then we see that the product
would belong to the ideal . But this is impossible because is an element of the multiplicative semigroup and does not intersect . Thus we can conclude that either or belongs to the ideal .
Let be prime, then since otherwise would be equal to . Now by theorem 2 is a cancellation ring. The canonical projection is a homomorphism, so is the identity element of . This in turn implies that the semigroup is a monoid with identity element . ∎
|Title||characterization of prime ideals|
|Date of creation||2013-03-22 15:22:01|
|Last modified on||2013-03-22 15:22:01|
|Last modified by||GrafZahl (9234)|
|Synonym||characterisation of prime ideals|