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A ring is a set R, together with two binary operations $+, \cdot : R\times R \to R$ that satisfy

  1. $(R,+)$ is an abelian group
  2. Associativity
  3. Distributivity
  4. Multiplicative identity so $\exists 1_R \in R$ such that $1a = a1 = a\, \, \forall \, a \in R$

Does this mean that the group is only closed under the '$+$' opearation and not the '$\cdot$' one?

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You wrote "the group" instead of "the set" in the formulation of your question, so let me just remind you that $(R,\cdot)$ is in general not a group. –  Per May 5 '12 at 13:19
    
Which group axiom does it not satisfy? –  user26069 May 5 '12 at 14:18
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Rings don't satisfy the axiom of inverses: What is the inverse of 0? [And this can be the case for other elements: Consider the integers. Does 2 have an (integer) inverse?] –  Alastair Litterick May 5 '12 at 14:21
    
"Associativity" by itself doesn't mean anything. You need to say "$+$ is associative and $\times$ is associative", or words to that effect. "Distributivity" by itself doesn't mean anything, and you need to specify who distributes over what. Here, we need "$\times$ distributes over $+$." –  Arturo Magidin May 5 '12 at 21:53

2 Answers 2

up vote 11 down vote accepted

The very definition of 'binary operation' implies closure: A binary operation is a map from $R \times R$ to $R$.

The only time you ever get problems with closure is when you have to define substructures (i.e. subrings, subgroups, etc.), when the target of the binary operation is potentially bigger than the source. In that context, yes, you require a subring to be closed under both binary operations.

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Closed means that whatever starts in R ends up in R. The $\cdot$ operation takes two elements in $R$ and produces another one in $R$, hence $R$ is closed under this operation (as well as with $+$).

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