Talk:Radical of an ideal

From Wikipedia, the free encyclopedia

This seems quite an odd change. Where do you define the radical of an arbitrary ideal? Not every radical is NilRadical, which is the radical of 0, not of anything else... mousomer

new definition : Let a be an element of a ring R. Then a is right quasi regular (r.q. r.) if there exists an element b of R such that a + b + ab = 0 , the more common definition , (a + b - ab = 0 , the definition of N. McCoy). Then b is a right quasi inverse of a. Left quasi regularity is similarly defined. A right ideal B is right quasi regular (r. q. r.) if all its elements are right quasi regular. There is a similar definition for a left quasi regular right ( or left) ideal. The Jacobson radical is J(R), the set a of elements of R such that the right ideal aR is r. q. r. for any ring R.

        -- S. A. G.

Retrieved from "http://en.wikipedia.org/wiki/Radical_of_an_ideal"

The article is still mixed up. Is this "radical of an Ideal" essay of "radical of a ring" essay? I think we should begin by giving alternate definitions of the radical of an Ideal (NOT OF A RING), and the giving the special case of the nilradical of a ring. Only after that, will it make sense to give the correspondence between the two notion. Any objections? mousomer 13:02, 28 September 2005 (UTC)

[edit] Confusion

The article nilpotent states that

Every nilpotent element in a commutative ring is contained in every prime ideal of that ring, and in fact the intersection of all these prime ideals is equal to the nilradical.

This article states that

If P is a prime ideal, then R/P is an integral domain, so it cannot have zero divisors, and in particular it cannot have nilpotents. Hence all prime ideals are radical.

Clearly, the article it as used here is a bit misleading: does "it" refer to "P" or to "R/P"? The "hence" doesn't seem to follow from what came before. I am loathe make any changes of wording here. linas 05:16, 29 December 2005 (UTC)

I reworded the article by adding more details. Wonder what you think. Oleg Alexandrov (talk) 16:02, 29 December 2005 (UTC)

Yes, thank you, that provided the missing link. linas 17:08, 29 December 2005 (UTC)