> 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.

Aaron Meurer

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