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.