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nil is a radical property
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(Proof)
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We must show that the nil property,
 , is a radical property, that is that it satisfies the following conditions:
- The class of
 -rings is closed under homomorphic images.
- Every ring
 has a largest
 -ideal, which contains all other
 -ideals of  . This ideal is written
 .
-
 .
It is easy to see that the homomorphic image of a nil ring is nil, for if
 is a homomorphism and  , then
 .
The sum of all nil ideals is nil (see proof here), so this sum is the largest nil ideal in the ring.
Finally, if  is the largest nil ideal in  , and  is an ideal of  containing  such that  is nil, then  is also nil (see proof here). So
 by definition of  . Thus  contains no nil ideals.
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"nil is a radical property" is owned by mclase.
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(view preamble)
Cross-references: proof, nil ideals, sum, homomorphism, nil ring, easy to see, ideal, contains, ring, homomorphic images, closed under, class, radical property, property, nil
There is 1 reference to this entry.
This is version 2 of nil is a radical property, born on 2004-02-28, modified 2004-02-28.
Object id is 5651, canonical name is NilIsARadicalProperty.
Accessed 2007 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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