# $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(Ae,N), N\otimes_{eAe}A=N\otimes_{eAe}eA$.

Let $e$ be an idempotent of a ring $A$ and $N$ is an $A$-module. Why $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(Ae,N), N\otimes_{eAe}A=N\otimes_{eAe}eA$? Can you prove this explicitly? Is the following true:

• $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(eA,N)$
• $N\otimes_{eAe}A=N\otimes_{eAe}Ae$
• $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(A,eN)$
• $N\otimes_{eAe}A=Ne\otimes_{eAe}A$
• $\mathrm{Hom}_{eAe}(A,N)=\mathrm{Hom}_{eAe}(A,Ne)$
• $N\otimes_{eAe}A=eN\otimes_{eAe}A$

Thank you.

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hint: $A = eA \oplus (1-e)A$ as right $A$-modules, but also as left $eAe$-modules. $eAe$ acts as zero on the second summand. This should help you write down the isomorphisms.