I'm studying Abstract Algebra right now, currently covering rings. In the introduction of the subject, I am curious as to why there is no need for there to be a multiplicative identity. I understand that in order to be a ring, we require the set to be an abelian group under addition operation and a monoid under multiplication. But what is the reason for the monoid, rather than group under multiplication--or lack of multiplication?
Definition of Ring
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$\begingroup$
abstract-algebra
ring-theory
definition
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6You seem to be asking two questions - are you asking about the multiplicative identity or multiplicative inverses? For the former, see http://math.stackexchange.com/questions/48587/definition-of-ring-vs-rng . For the latter, well, that's what fields are for. – 2011-07-24
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3Related questions [here](http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity) and [here](http://math.stackexchange.com/questions/48587/definition-of-ring-vs-rng). – 2011-07-24
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0It's the former. Thanks both for the links. – 2011-07-24
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0Also: http://mathoverflow.net/questions/22579/ – 2011-07-24
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0See also this question on [examples/motivation for non-unital rings.](http://math.stackexchange.com/questions/37705/non-unital-rings-a-few-examples/37716#37716) Note that the definition of a ring also requires that the additive and multiplicative structures are related - by the distributive law. Without such one would simply have a set with two completely unrelated structures. – 2011-07-25
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0Regarding the statement of the question: if the ring has a multiplicative identity then it is a monoid under multiplication. If a multiplicative identity does not exist then the ring (also called general ring) is not a monoid under multiplication. – 2011-07-25