A categorical way of viewing : If $M$ over $R$ is free then $M/IM$ over $R/I$ is free

Question

Let $R$ be a commutative ring with unity , let $I$ be an ideal of $R$ and $M$ be a module over $R$ . I know that if $M$ over $R$ is a free module , then $M/IM$ is a free $R/I$ module . I want to know ; is there any categorical way to see this ? I know that for fixed $R$ and $I$ , from the category of $R$ modules to the category category of $R/I$ modules , there is a functor sending $M$ to $M/IM$ . Can we see the freeness of $M/IM$ , given the freeness of $M$ , in a similar way ?

Answer 1

The categorical fact here is that you can "compose adjunctions", in the sense that if you have functors $F_1:\mathcal{X}\to \mathcal{Y}$, $F_2:\mathcal{Y}\to \mathcal{Z}$, $G_1:\mathcal{Y}\to \mathcal{X}$ and $G_2:\mathcal{Z}\to \mathcal{Y}$ such that $F_1\dashv G_1$ and $F_2\dashv G_2$, then $F_2\circ F_1\dashv G_1\circ G_2$.

Indeed, the construction of free modules gives a functor $F_R:\mathbf{Set}\to R-\mathbf{Mod}$, which is left adjoint to the forgetful functor $U_R: R-\mathbf{Mod}\to\mathbf{Set}$; and since this holds for any ring, we also have an adjunction $F_{R/I}\dashv U_{R/I}$ between $\mathbf{Set}$ and $R/I-\mathbf{Mod}$.

Now the functor $M\mapsto M/IM$ is isomorphic to the functor $\pi_!: M\mapsto R/I\otimes_{R}M$ (see for example this question, although it doesn't adress the naturality), which is left adjoint to the functor $\pi^*:R/I-\mathbf{Mod}\to R-\mathbf{Mod}$ induced by the quotient $\pi:R\to R/I$ (this is known as extension of scalars or change of rings). Since this functor does not change the underlying set of the module, we have $U_{R/I}=U_{R}\circ \pi^*$.

Now the composition of adjunctions I mentioned above and the uniqueness of adjoints tells you that $F_{R/I}=\pi_!\circ F_{R}$; thus if $M$ is free over a set $X$ (i.e. if $M=F_R(X)$), then $M/IM$ is free over the same set $X$, because $M/IM=\pi_!(F_R(X))=F_{R/I}(X)$.