Is a ring closed under both operations?

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?

• 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
• 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

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