characterize quotient objects of a ring We know that the quotient objects of an $A-module$ $R$ are except equivalence, the quotient modules $\frac{R}{S}$ with the homomorphisms $h: R\to \frac{R}{S}$ , $h(r) = r + S$ with $r\in R$.
But characterizing the quotient objects of a ring does not seem to be that simple. It seems to be because in the category of rings the epimorphisms do not coincide with the surjective homomorphisms. My question Can we characterize the quotient objects of a ring??
 A: You implicitly identify "quotient object" with "epimorphism." It's not at all clear that this is a useful definition. It's right there in the name: a quotient object should be an object given as a quotient in some way, and not all epimorphisms have this property. (In other words, "quotient object" is not perfectly categorically dual to "subobject"! Only in very nice categories like abelian categories is that true.)
A useful and general notion of "quotient object" in a category is the following: any morphism $f : a \to b$ has a kernel pair $a \times_b a$ (in $\text{Set}$ this computes the equivalence relation on $a$ induced by $f$), and we say that $f$ is an effective epimorphism if $f$ is the coequalizer of its kernel pair (in $\text{Set}$ this computes the quotient by the equivalence relation induced by $f$).
This is a "nonlinear" generalization of "$f$ is the cokernel of its kernel" and you can check that for rings, the effective epimorphisms are precisely the surjections, as follows. If $f : R \to S$ is a morphism its kernel pair is the congruence (internal equivalence relation)
$$R \times_S R = \{ (r_1, r_2) \in R^2 : f(r_1) = f(r_2) \}$$
which is exactly $\{ (r_1, r_2) \in R^2 : r_1 - r_2 \in I \}$ where $I = \text{ker}(f)$. Now you can check that the coequalizer of the two projections $R \times_S R \rightrightarrows R$ computes the quotient $R/I$.
See also this blog post. A nice result you can prove is that a morphism is an isomorphism iff it's a monomorphism and an effective epimorphism; note that this is false if "effective" is dropped.
