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.
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.
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