## Incorrect Statement in Proof

Since S is multiplicatively closed and R has identity, Zorn's lemma says that there exists an ideal P in this ring which contains I and is maximal with respect to being disjoint from S.

The application of Zorn's Lemma does not use the multiplicative closedness of R. This is used in establishing the primality of the ideal P.

If a proof is included, it should be correct. One could also delete the proof since usually only summary statements are included.

I see what you mean, and I think the problem occurred when a correctly summarized statement was turned into a proof by later editors. The multiplicative property alone is not enough to show that the ideal is prime (we need the maximality too) but I'll go ahead and do that myself. Rschwieb (talk) 14:20, 27 December 2012 (UTC)
I think the problem is the paragraph is doing too many things at once. The key point used here is the so-called multiplicative avoidance, that is, an ideal disjoint from a multiplicative set S may be enlarged to a prime ideal still being disjoint from S. This is a typical combination of Zorn's lemma and other properties. This is a standard argument and makes sense to include it in Wikipedia. (in any case a new version looks fine?) -- Taku (talk) 12:14, 28 December 2012 (UTC) and basically this comes down to a "well-known" tendency of a maximal element to become a prime ideal. -- Taku (talk) 12:22, 28 December 2012 (UTC)
The Lam-Reyes papers on that phenomenon are among the most fascinating I've ever read. I highly recommend them. Rschwieb (talk) 14:43, 28 December 2012 (UTC)
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## Odd change

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.


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)

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## 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.

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)
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## Notation

Hi, I like the notation r for radicals more than the other ones like sqrt. Is there any objection to the notational change? Atiyah-MacDonald uses r too. (don't know about Eisenbud). -- Taku (talk) 21:15, 5 November 2012 (UTC)

Hmm, I've only ever seen \sqrt and rad(). I've only seen r used for annihilators. Rschwieb (talk) 14:51, 6 November 2012 (UTC)
Eisenbud uses "rad" while Bourbaki uses $\mathfrak{r}$. Matsumura uses sqrt; can't find any uniformity :) I think rad() is not a good choice since it could refer to the radical of a ring, which is a completely different concept. -- Taku (talk) 20:24, 6 November 2012 (UTC)
I agree that "rad" seems too vague. The radical symbol on the other hand is not being competed for, and really does invoke the imagery of containing "roots" of things in the idea. Rschwieb (talk) 14:18, 7 November 2012 (UTC)
I just noticed that our Hilbert's Nullstellensatz uses sqrt too. It makes sense to use the consistent notation. I guess we have the winner (I'm making change in a few days.) -- Taku (talk) 17:22, 7 November 2012 (UTC)
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