Theorem 1 (Lasker-Noether).
The theorem can be proved in two steps:
Let be the set of all ideals of a Noetherian ring which can not be written as a finite intersection of irreducible ideals. Suppose . Then any chain in must terminate in a finite number of steps, as is Noetherian. Say is the maxmimal element of this chain. Since , itself can not be irreducible, so that where and are ideals strictly containing . Now, if , then then would not be maximal in the chain . Therefore, . Similarly, . By the definition of , and are both finite intersections of irreducible ideals. But this would imply that , a contradiction. So and we are done. ∎
Every irreducible ideal in is primary
Suppose is irreducible and . We want to show that either , or some power of is in . Define , the quotient of ideals and . Since
we have, by one of the rules on quotients of ideals, an ascending chain of ideals
Since is Noetherian, for all . Next, define , the sum of ideals and . We want to show that .
First, it is clear that and , which takes care of one of the inclusions. Now, suppose . Then , where and , and . So, . Now, , so . But this means that , and this proves the other inclusion.
A ring is said to be Lasker if every ideal is decomposable. The theorem above says that every commutative Noetherian ring with 1 is Lasker. There are Lasker rings that are not Noetherian.
|Date of creation||2013-03-22 18:19:53|
|Last modified on||2013-03-22 18:19:53|
|Last modified by||CWoo (3771)|