I am solving an exercise from Dummit, Foote. Let $R$ be a ring with $1$. Then the following are equivalent:
Every $R$-module is projective.
Every $R$-module is injective.
(1) implies that for any R-mod $A$ we have $Ext^i(-,A)$ is a zero functor. But this means that for any R-mod $Ext^i(L,A) = 0$ for all $A$ so $L$ is injective. I show $(2) \Rightarrow (1)$ exactly the same way.
It is quite easy so I am afraid I missed something. Is my proof correct?