Talk:Ring theory
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[edit] comment by Tarquin
- Euclidean domain => Principal ideal domain => unique factorization domain => integral domain => Commutative ring.
Is the above a hierarchy of inclusion? If so, use the subset symbol. -- Tarquin
- It's inclusion, but it doesn't make much sense to use subset symbol, because there aren't any standard "symbols" for "the class of all Euclidean domains", e.g., unless you want to make up several just for this article. Writing "Euclidean domain contained in PID contained in UFD, etc." is misleading, because it makes it sound like a ED is set-theoretically contained in a PID, contained in a UFD, not the same. Revolver 02:08, 11 Jun 2004 (UTC)
A ring is called commutative if its multiplication is commutative. The theory of commutative rings resembles the theory of numbers in several respects, and various definitions for commutative rings are designed to recover properties known from the integers. Commutative rings are also important in algebraic geometry
- Except that the ring article contradicts this and requires commutativity, which adds credence to my belief that this shouldn't be required in the definition. Revolver 02:08, 11 Jun 2004 (UTC)
- No, multiplication in rings is not generally required to be commutative. I double-checked this in Herstein's Topics in Algebra as a sanity check. Isomorphic 02:19, 11 Jun 2004 (UTC)
- I was wrong, the ring article doesn't require it, it requires unity. My fault. (Although I disagree with unity requirement, as well, that's another matter.) Actually, I just disagree with these universal wikipedia definitions (rather than article by article basis).
- There are important examples of rings that do not have a identity. Regardless, and identity can always be formally adjoined by using the adjoint of the forgetful functor from the category of rings with unity to the category of rings.
- I was wrong, the ring article doesn't require it, it requires unity. My fault. (Although I disagree with unity requirement, as well, that's another matter.) Actually, I just disagree with these universal wikipedia definitions (rather than article by article basis).
- No, multiplication in rings is not generally required to be commutative. I double-checked this in Herstein's Topics in Algebra as a sanity check. Isomorphic 02:19, 11 Jun 2004 (UTC)
[edit] Vote for new external link
Here is my site with ring theory example problems. Someone please put this link in the external links section if you think it's helpful and relevant. Tbsmith
http://www.exampleproblems.com/wiki/index.php/Abstract_Algebra#Rings
[edit] Patent nonsense
Sorry for exaggerating in my edit comment -- the patent nonsense was only in Wikipedia for a little over a day before I reverted it. (I misread the date.) I find it embarrassing, though, that someone who trusts Wikipedia asked me for an explanation of it.—GraemeMcRaetalk 04:48, 21 January 2006 (UTC)