nilpotency is not a radical property
Let be a field, and let be the ring of polynomials over in infinitely many variables . Let be the ideal of generated by . Let . Note that is commutative.
For each , let . Let .
Then each is nilpotent, since it is the sum of finitely many nilpotent ideals (see proof http://planetmath.org/node/5650here). But is nil, but not nilpotent. Indeed, for any , there is an element such that , namely , and so we cannot have .
So cannot have a largest nilpotent ideal, for this ideal would have to contain all the ideals and therefore .
|Title||nilpotency is not a radical property|
|Date of creation||2013-03-22 14:13:02|
|Last modified on||2013-03-22 14:13:02|
|Last modified by||mclase (549)|