Xantharius

Course structure

Created the article coarse structure today: please help with formatting and so on if you like. Xantharius 20:11, 1 May 2006 (UTC)

Clout Fantasy

Hi, can we talk about how to get the "non-neutral content marker" off the Clout Fantasy page? I've re-edited it several times to attempt to make it neutral and would like to know if I've succeeded. I"m not sure how to proceed from here. Thanks, Jirel.

Generational secrets

Thanks dear! Might have to slug a couple of whiskey neat, after this little battle over Generational secrets. I think I've bent a few minds. To the "Revelator" and your Winged Self, Pax et Agape, Tatee Estelle

Joe Bray

Thanks for the tip. I must have forgotten to plug that in there. Joeraybray 03:06, 18 July 2007 (UTC)

Um...

My edit said that Reals were not closed, so it's not clear why you reverted it.- (User) WolfKeeper (Talk) 01:40, 2 March 2008 (UTC)

Hello and thanks

Hello, Wiki fairy and thanks for your useful work. This is for the article Eigenvalue, eigenvector, and eigenspace but you have done those embelishments to many more. Keep up the good work and don't think that no one notices. --Lantonov (talk) 17:15, 13 March 2008 (UTC)

Thanks for converting math formulas to HTML

However: [1]

-- Dominus (talk) 03:40, 16 March 2008 (UTC)

Zero-product property

I saw your edit, but isn't it too obvious to be worth mentioning? By definition, an integral domain is a commutative ring with unity without zero divisors. Saying that every (commutative rings with unity that is not an integral domain) has zero divisors is almost rephrasing the definition. If we drop the unity, then it's easy to construct a commutative ring without zero divisors that is not an integral domain: take, for example, the subring generated by x in the ring Z[x]. (I brag myself as The Overlord of Counter-examples :-) ). Albmont (talk) 12:47, 25 March 2008 (UTC)

Until a few days ago the article said that a ring which was not an integral domain had zero divisors, which is obviously untrue as the counter-example of the quaternions showed. I agree with you entirely that this is rephrasing the definition. But how else do we talk about rings which are not integral domains? Since the zero-product property is in fact the defining property of an integral domain, if we remove this then zero divisors must arise. I can't at the moment think of a better way of phrasing this which still highlights what happens when commutative rings fail to have the zero-product property. Xantharius (talk) 18:12, 25 March 2008 (UTC)

In fact, the integral domain is a ring with 3 independent axioms added to it: unit, commutativity and no zero divisors. And I just showed a commutative ring that is not an integral domain, that has no zero divisors: the ring of x p(x), where p(x) is a polynomial with integer coefficients. BTW, I think I can provide a quite pathological example of a commutative ring that is not an integral domain, has no zero divisors and can't be immersed into an integral domain <evil grin> Albmont (talk) 18:30, 25 March 2008 (UTC)

Well, it all depends on one's starting point (whether we are discussing commutative rings in the first place, for example) to determine what makes an integral domain an integral domain, but from the perspective of the zero-product property neither unity nor commutativity are the real focus: it's the zero-product property that counts, in my view, and I wonder whether it benefits the reader to make too much of a deal about this. Still, correctness is desirable.

Of course, any commutative ring R with unity with a rule for multiplication defined by ab = 0 for all a, b in R is an example of a commutative ring with unity which is not an integral domain, because everything is a zero divisor. But what's your example? Xantharius (talk) 19:47, 25 March 2008 (UTC)

Your example is not a commutative ring with unity: there's no unity, because 1.a = 0. If a commutative ring with no zero divisors can be expanded to an integral domain, then it can be immersed in the field of quotients of that domain, so this suggests the construction of the pathological example: it's enough to have one where $x^{n}=a\,$ has n+1 (or more) solutions. But I can't find a simple enough counter-example. Albmont (talk) 20:16, 25 March 2008 (UTC)

Oops. I think I can't build such counterexample. I am almost convincing myself of the opposite: every commutative ring with the zero-product property can be immersed in a field. Too bad, that's one class of counterexamples that vanish :-( Of course if x^2 = a has three distinct roots, then there are zero divisors. It's quite easy to prove it for x^3 = a, and probably there's a trivial induction to extend to n. Albmont (talk) 20:29, 25 March 2008 (UTC)

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Intermediate value theorem

The proof seems to be missing some formulas. Katzmik (talk) 12:23, 2 September 2008 (UTC)

D&D articles for Wikipedia 0.7

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File:Zerodivisor.png listed for deletion

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