# Composition of module homomorphisms and their kernels

Let $R$ be a principal ideal ring and let $I$ and $J$ be two ideals of $R$. Suppose $\phi: R \times R \rightarrow R \times R$ and $\psi: R \times R \rightarrow R \times R$ are two $R$-module homomorphisms. Suppose further that $\ker(\phi) = I \times <0>$ and $\ker(\psi) = <0> \times J$.

Then $\psi \circ \phi$ is an $R$-module homomorphism defined on $R \times R$. What can we say about the kernel of $\psi \circ \phi$? Under what conditions will it happen than $\ker(\psi \circ \phi) = I \times J$? Will this be dependent on the map? Any over-arching theorems dealing with this sort of thing?

I have a similar type question here. Any sources other than Dummit and Foote would be good too. (They spend too much time on modules over PID's which doesn't help me.)

• Two hints: (i) $\ker(\psi \circ \phi) = \phi^{-1}[\ker(\psi)]$; (ii) linking back to a trail of previously unanswered questions is not likely to be productive on MSE - think of the amount of reading you are requiring people to do. If you want answers to the earlier questions, go back and try to simplify or clarify the questions or offer a bounty. – Rob Arthan Aug 25 '15 at 20:30
• I don't have bounty privileges but I will look at the inverse map on the kernels. That's a great idea! Thanks!!! – Greg Doyle Aug 25 '15 at 20:32
• I thought 50 points was enough. Anyway, you have a few more reputation points now. $\ddot{\smile}$ – Rob Arthan Aug 25 '15 at 20:35
• Thanks for the points! A problem though, since $\phi$ and $\psi$ are homomorphisms, an inverse map doesn't necessarily exist. – Greg Doyle Aug 25 '15 at 20:45
• @GregDoyle $\phi^{-1}(\ker(\psi))$ is the set of elements $(x,y)\in R\times R$ such that $\phi(x,y)\in \ker(\psi)$; it's called inverse image or preimage of a set, and it is well-defined even if $\phi$ is not bijective. – Arnaud D. Aug 25 '15 at 21:02