# A ring isomorphic to an indecomposable ring is also indecomposable

Recall that a ring $$R$$ with unity is called indecomposable if $$R \neq 0$$ and $$R$$ can not be expressed as a direct product $$R= R_1 \times R_2$$ for non-zero rings $$R_1$$ and $$R_2$$.

I want to prove this claim: If $$f:R\to S$$ is an isomorphism of rings and $$R$$ is indecomposable, then $$S$$ is also indecomposable.

Here is my attempt: Suppose, seeking a contradiction, that $$S$$ is decomposable as $$S=S_1\times S_2$$, where $$S_1$$ and $$S_2$$ are nonzero rings. I need a contradiction by showing that $$R=R_1\times R_2$$ for some two nonzero rings $$R_1$$ and $$R_2$$. My question is how to get such rings $$R_1$$ and $$R_2$$.

Another question is: Is the definition given above of indecomposable rings correct?

Note: I need to prove the claim by using the definition only.

• Your definition is incorrect. The correct definition is that a ring is indecomposable if and only if it is not isomorphic to a product of nonzero rings. Whenever you’re dealing with the equality of two rings, you’re probably on the wrong track with your definition. Nov 20, 2021 at 23:37
• Thanks a lot @MarkSaving . However, this is not my own definition. This is a definition found in an e-book. Nov 20, 2021 at 23:42
• @MarkSaving I see no problem writing it as an equality. It is common practice to elide the isomorphism because, well, the difference between isomorphic objects is not often relevant to an algebraist. Nov 21, 2021 at 3:06
• @rschweib This is a good philosophy. Of course, once you do this, everything is automatically isomorphism invariant (as it should be). Nov 21, 2021 at 3:13
• Of course. I just wanted to point out that the matter of eliding isomorphisms is rather subtle and it is difficult even for me, a (now out-of-practice) logician and category theorist, to formulate and articulate the principle. (The best I can do is this: if you have two distinct objects and one isomorphism between them, you are entitled to identify them and replace the isomorphism by the identity map, but then from that point the formerly distinct objects are identical so you are no longer allowed to replace arbitrary isomorphisms between them with the identity map.) Nov 21, 2021 at 12:30

The property of being directly decomposable is equivalent to the existence of a nontrivial central identity $$e$$ (meaning that $$e^2=e$$ and $$e$$ is not the additive or multiplicative identity) which is the identity of one of the rings (the other identity has to be $$1-e)$$.
OK, well, I think I see that the definition given and the question given in conjunction do indeed warrant more detail. Let's pedantically insist that equality has to mean set equality and nothing more. Then $$S\cong R_1\times R_2$$ via an isomorphism. Using the images of $$R_1$$ and $$R_2$$ via the isomorphism, one finds subsets of $$S$$ (ideals actually) such that $$S=f(R_1\times \{0\})\oplus f(\{0\}\times R_2)$$. Since $$S=I\oplus J$$ can be exchanged for $$S=I\times J$$ by recognizing the internal direct sum as a product, you would have that every decomposition of $$R$$ (in the rigid sense) yields a corresponding decomposition of $$S$$ (in the rigid sense.) One is indecomposable iff the other one is.
• Did you notice the statement "Note: I need to prove the claim by using the definition only." written in my question?. I need two things: 1) A correction of the definition written by me above if it's incorrect (In other words, which is more accurate? writing $R\cong S\times T$ or $R=S\times T$) and 2) a proof using ONLY the definition not using an equivalent criterion. Thanks in advance. Nov 21, 2021 at 10:46
• How can one recognize the internal direct sum as a product?!! It's known that the internal direct sum is ISOMORPHIC to the external direct sum and the finite external direct sum and the direct product coincide. That is, one must write $R\oplus S \cong R\times S$ not $R\oplus S = R\times S$. Nov 22, 2021 at 3:53