Effect of integral domain structure on the additive group of the ring I was wondering about the implications of the following statement:
The characteristic of an integral domain $R$ must be zero or prime $p$.
This implies that all elements in $R\backslash\{0\}$ should have order $p$ in the commutative group $R,+$. So imposing the ring structure without zero divisors on a commutative group $R,+$ might result in a drastic change of the group structure, because suddenly all the elements are of the same order $p$? How can this be understood intuitively?
 A: By "imposing a ring structure" on $\langle R,+\rangle$ I take that you mean that first one has an abelian group $\langle A,+,-,0\rangle$ and then one considers a distributive product $*:A\times A\to A$.  If $R=\langle A,*,+,-,0\rangle$ satsifies the axiom of a ring then you might say you have "imposed" a ring structure on $A$.
In that case your question could be formalized as: what conditions on $A$ permit the construction of a ring $R$ that is an integral domain of characteristic $p$?
The answer as you noted is $A\cong \bigoplus_{i\in I} \mathbb{Z}_p$ (subject to axiom of choice).  In particular $A$ is a vector spaces.
In fact that would be the natural proof.  Suppose $R$ is a ring of characteristic $p$.  Then the subring $C=\{0,1,\ldots,p-1\}$ acts on $R$.  But $C$ is a field.  So $R$ is a $C$-vector space.  Then, under the axiom of choice this vector space has a basis and so $R\cong \bigoplus_{i\in I} \mathbb{Z}/p$.
(Edited to include @Max's valid point.  When a basis is infinite we still mean linear combinations to be finite support; hence, I should not have written $\prod$.)
