existence of maximal ideals
Every unital ring has a maximal ideal.
Proof of theorem. This proof is a straightforward application of Zorn’s Lemma. Readers are encouraged to attempt the proof themselves before reading the details below.
Note that , so is non-empty.
In order to apply Zorn’s Lemma we need to prove that every non-empty chain (http://planetmath.org/TotalOrder) in has an upper bound in . Let be a non-empty chain of ideals in , so for all indices we have
We claim that defined by
is a suitable upper bound.
is an ideal. Indeed, let , so there exist such that , . Since these two ideals are in a totally ordered chain we have
Without loss of generality, we assume . Then both , and is an ideal of the ring . Thus .
Similarly, let and . As above, there exists such that . Since is an ideal we have
Therefore, is an ideal.
, otherwise would belong to , so there would be an such that so . But this is impossible because we assumed for all indices .
. Indeed, the chain is non-empty, so there is some in the chain, and we have .
Therefore . Hence every chain in has an upper bound in and we can apply Zorn’s Lemma to deduce the existence of , a maximal element (with respect to inclusion) in . By definition of the set , this must be a maximal ideal of containing . QED
Note that the above proof never makes use of the associativity of ring multiplication, and the result therefore holds also in non-associative rings. The result cannot, however, be generalized to rings without unity.
Note also that the use of the Axiom of Choice (in the form of Zorn’s Lemma) is necessary, as there are models of ZF in which the above theorem and corollary fail.
|Title||existence of maximal ideals|
|Date of creation||2013-03-22 13:56:57|
|Last modified on||2013-03-22 13:56:57|
|Last modified by||yark (2760)|
|Synonym||existence of maximal ideals|