On Thu, Apr 5, 2012 at 1:13 AM, Aaron Meurer <[email protected]> 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), > 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.
Hm, I see. The page [0] says: Many authors do not require rings to have a multiplicative identity,[1] so the concept discussed here is just what these authors call a ring. Apparently, the professor who taught me ring theory belong to exactly this camp. However, as I can see, there are slightly more people who define ring as having the multiplicative identity; in this case I agree with the fact that Ideal should not derive from Ring. Sergiu [0] http://en.wikipedia.org/wiki/Pseudo-ring -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
