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nilpotency is not a radical property
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(Proof)
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Nilpotency is not a radical property, because a ring does not, in general, contain a largest nilpotent ideal.
Let  be a field, and let
![$ S = k[X_1, X_2, \dotsc]$](http://images.planetmath.org:8080/cache/objects/5652/l2h/img2.png "$ S = k[X_1, X_2, \dotsc]$") 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 here). 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  .
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"nilpotency is not a radical property" is owned by mclase.
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(view preamble)
Cross-references: nil, proof, sum, nilpotent, commutative, generated by, ideal, variables, polynomials, field, nilpotent ideal, contain, ring
There is 1 reference to this entry.
This is version 1 of nilpotency is not a radical property, born on 2004-02-28.
Object id is 5652, canonical name is NilpotencyIsNotARadicalProperty.
Accessed 1430 times total.
Classification:
| AMS MSC: | 16N40 (Associative rings and algebras :: Radicals and radical properties of rings :: Nil and nilpotent radicals, sets, ideals, rings) |
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Pending Errata and Addenda
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