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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?

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up vote 2 down vote accepted

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?

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Thank you,I get this question in the representation theory for quiver.If Q is a quiver,then KQ is indecomposable iff Q is connect iff KQ is prime.If the ring R has the form KQ(for example.some matrix ring whose element is a power of field),the R is indecomposable iff R is prime.But I want to know some more. – Strongart May 1 '12 at 11:06
What type of quivers on which type of rings are you working with? – rschwieb May 1 '12 at 12:35
I study quiver less than two months,just know some common situation – Strongart May 4 '12 at 11:50
What type of quivers on what type of rings? – rschwieb May 4 '12 at 13:24
The common situation is in the Assem's book:Elements of the Representation Theory of Associative Algebras, Corollary 1.11. – Strongart May 7 '12 at 11:42

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