subdirectly irreducible ring
Equivalently, a ring is subdirectly irreducible iff the intersection of all non-zero ideals of is non-zero.
Let be the set of all non-zero ideals of .
. Suppose first that is subdirectly irreducible. If , then is a subdirect product of , for given by is injective. If , then for all , or , or . But then given by is not an isomorphism for any , contradicting the fact that is subdirectly irreducible. Therefore, .
. Suppose next that . Let be a subdirect product of some , and let . Each is an ideal of . Let . If is not an isomorphism (therefore not injective), is non-zero. This means that if is not subdirectly irreducible, . But , contradicting the subdirect irreducibility of . As a result, some , or is an isomorphism. ∎
As an application of the above equivalence, we have that a simple ring is subdirectly irreducible. In addition, a commutative subdirectly irreducible reduced ring is a field. To see this, let be the set of all non-zero ideals of a commutative subdirectly irreducible reduced ring , and let . So by subdirect irreducibility. Pick . Then . So since is minimal. This means , or , which means . Now, let any , then , so for some , which means is a field.
|Title||subdirectly irreducible ring|
|Date of creation||2013-03-22 14:19:13|
|Last modified on||2013-03-22 14:19:13|
|Last modified by||CWoo (3771)|