For every ring R, every left R-module M can be imbedded as a submodule of an injective left R-module

Question

Prove that for every ring $R$, every left $R$-module $M$ can be imbedded as a submodule of an injective left $R$-module

Answer 1

There is a standard dualization trick : for any $R$-module $A$, write $A^* = Hom_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z})$, which is a $R$-module. This gives a contravariant functor from the category of $R$-modules to itself.

Then write $A^*$ as quotient of a free module : $\bigoplus_{i\in I} R \to A^*\to 0$. Dualizing : $0\to A^{**}\to \prod_{i\in I} R^*$ (this is exact because $\mathbb{Q}/\mathbb{Z}$ is an injective $\mathbb{Z}$-module, since it's divisible).

Now $R^*$ is injective (because $Hom_\mathbb{Z}(R,\bullet)$ has a left adjoint and $\mathbb{Q}/\mathbb{Z}$ is injective, so it's sent to an injective object), so $\prod_{i\in I} R^*$ is injective.

We conclude by the fact that the canonical $A\to A^{**}$ given by $a\mapsto ev_a$ (the evaluation map at $a$) is injective, so $A\to A^{**}\to \prod_{i\in I} R^*$ is an injective morphism to an injective $R$-module.