Proving an $R$-module isomorphism is in fact an $R/I$-module isomorphism using universal properties.

Question

Notation. Given a commutative ring $R$ and a set $S$, write $F_R(S)$ for the free $R$-module on $S$.

Here's a classical exercise:

1. Given an ideal $I\vartriangleleft R$, prove $\frac{F_R(S)}{I\;F_R(S)}\cong F_{\frac{R}{I}}(S)$.
2. Deduce that free modules over commutative rings have IBN.

The first part can be proved by constructing an $\frac{R}{I}$-linear surjection $F_{\frac{R}{I}}(S)\to \frac{F_R(S)}{I\;F_R(S)}$ using the universal property and then proving it's injective.

For the second part, I am tempted to consider the following diagram in the category of $R$-modules, where an isomorphism $F_R(S)\cong F_R(S^\prime)$ induces the rest of the horizontal isomorphisms. The problem is that the induced isomorphism $\frac{F_R(S)}{I\;F_R(S)}\cong \frac{F_R(S^\prime)}{I\;F_R(S^\prime)}$ is apriori only $R$-linear as opposed to $\frac{R}{I}$-linear, and the latter is needed to complete the exercise.

$$\require{AMScd} \begin{CD} I\; F_R(S) @>>> I\; F_R(S^\prime)\\ @VVV @VVV\\ F_R(S) @>>> F_R(S^\prime) \\ @VVV @VVV\\ \frac{F_R(S)}{I\;F_R(S)} @>>> \frac{F_R(S^\prime)}{I\;F_R(S^\prime)} \\ @VVV @VVV\\ F_{\frac{R}{I}}(S) @>>> F_{\frac{R}{I}}(S^\prime) \end{CD}$$

Question. What is the right way to deduce the $R/I$-linearity of the induced $\frac{F_R(S)}{I\;F_R(S)}\cong \frac{F_R(S^\prime)}{I\;F_R(S^\prime)}$ from the universal property of the quotient ring $R/I$?

Answer 1

Here's another way to prove that if $f:R\to S$ is a ring epimorphism, then scalar restriction $f^\ast$ is full. I think it is a bit more structural.

Let $X,Y$ be two $S$-modules and write $f^\ast X,f^\ast Y$ for their pulled back $R$-module structures. We want to prove any $R$-linear map $\varphi:f^\ast X\to f^\ast Y$ satisfies $s\varphi(x)=\varphi (sx)$. We shall do this universally by considering all such morphisms at once.

Recall ${}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y)$ is an abelian group. As any abelian group, we may consider its endomorphism ring.

This endomorphism ring admits two ring morphisms from $S$, $$S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$ with one given by $s\mapsto (\varphi(x)\mapsto s \varphi(x))$ and the other by $s\mapsto (\varphi(x) \mapsto \varphi (sx))$. These are ring morphisms because $f^\ast X,f^\ast Y$ are $S$-modules. We want to show these two ring morphisms coincide.

I claim these ring morphisms coincide upon precomposing with $f:R\to S$.

$$R\to S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$ Indeed this amounts to the equation $f(r)\varphi(x)=\varphi(f(r)x)$ which holds because $X,Y$ are $S$-modules. We now apply the fact $f$ is epic to conclude the two of $S$-actions were already equal, as desired.