What is the relation between prime ring and indecomposable ring? Prime ring means $0$ is prime in the ring, so $R$ is a prime ring iff for any two elements $a$ and $b$ of $R$, $arb = 0$ for all $r\in R$ implies that either $a = 0$ or $b = 0$. Indecomposable ring means it is not the direct sum of nonzero ring.
I know that $\mathbb{Z}/4\mathbb{Z}$ is an indecomposable ring, but not prime because a commutative prime ring is a domain. But how about the opposite? And in which case they are the same?
 A: That's an OK definition of "prime" but I find another one (product of two right ideals is zero iff one of them is already zero) useful to keep in mind.
A prime ring is always indecomposable: if $R=S\oplus T$, then $S$ and $T$ are ideals of $R$ whose product is zero.
I can't offer any support for a sort of converse, because there does not seem to be a strong connection. There is not even a good result for commutative Artinian rings. A commutative Artinian ring is indecomposable iff local, but a commutative Artinian ring is prime iff it is a field.
Any local Artinian ring which is not a division ring is indecomposable (thanks to the absence of any idempotents except 0 and 1) but it isn't prime because its radical is a nonzero nilpotent ideal. This includes your example, which is Artinian and local. 
Let's say you wanted to show that nonprime rings with condition X are decomposable. You could reason: "There exist nonzero ideals $A$ and $B$ such that $AB=0$. Because of X there exists a nontrivial central idempotent of $R$." Maybe someone can think of a clever condition X?
