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.
Thanks in advance.
 A: 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 $1-e)$.
So ask yourself: do isomorphisms preserve nontrivial central idempotents?


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 ≅× or =×) and 2) a proof using ONLY the definition not using an equivalent criterion. –
Hussein Eid

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.
