On Wed, Apr 4, 2012 at 4:20 PM, Tom Bachmann <e_mc...@web.de> wrote: > On 04.04.2012 23:13, Aaron Meurer wrote: >>> >>> I'm not sure I see your point here; indeed, an ideal of ring with a >>> multiplicative unit need not necessarily include the unit, but this >>> doesn't make it less of a ring. >> >> >> Yes it does. It makes it not a ring at all. A ring is a set with two >> binary operations, + and *, *with the properties* that it is a group >> under each operation (is associative, closed, contains the identity), > > > Group implies inverses. 0 never has a multiplicative inverse unless we are > in the zero ring.
Ah yes, I forgot about that caveat (it's been a while I'm afraid). So what I said is not exactly true. My point is that if a set does not have 1, it's not a ring. > > >> is commutative under +, and has the distributive property (some >> definitions also require that 1 != 0, though I personally find this to >> be unnecessary). If it lacks any of these properties, it is not a >> ring. Saying something like "a ring with a multiplicative unit" is >> meaningless. Every ring has a multiplicative unit, by definition. >> > > Not quite. Every definition of ring I know requires a set with two binary > operations + and *, such that we get an abelian group under +, > distributivity, associativity for *. One may require the existence of a > multiplicative identity and then typically requires homomorphism to preserve > it (I think omitting this is sufficiently obscure to have invented the term > "rng"). One may require multiplication to be commutative. > > In my background "ring" is synonymous with commutative unital ring. As far > as I know there is quite a large body of mathematics (mainly related to > representation theory) devoted to the study of non-commutative rings as > well. I don't know anything interesting about non-unital rings (but most > books I know state that their rings are required to have identities, so I > guess its not completely out of the ordinary to omit them). > > All of this is really just nitpick for saying: while some in contexts rings > without identity are fine, we should not be required to treat ideals in this > way. So an ideal class structure would represent this. CommutativeRing would derive from Ring, but not all subclasses of Ring would be subclasses of CommutativeRing. It may make sense to have Ring derive from Ideal. If you remove important aspects like the multiplicative identity, it ceases to be a ring, which is why we have all these other lesser classes like Ideals. Aaron Meurer > > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to sympy@googlegroups.com. > To unsubscribe from this group, send email to > sympy+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sympy@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
