Finding a basis of $S \otimes_R N$ with $N$ a free $R$-module

Question

Let $S$ and $R$ be commutative rings with $f: R \to S$ a ring morphism. Suppose that $N$ is free $R$-module with basis $B$. I want to understand why $N \otimes_R S$ is a free $S$-module where the action is given by $s'\cdot (n\otimes s) = n \otimes s's$.

It's easy to see that $B$ generates $N \otimes_R S$, but I'm having trouble showing directly that $B$ is linearly independent over $S$. Perhaps there is an easy way to do this via the universal property of tensor products, but I'm unsure. Or perhaps I'm looking at the wrong set altogether.

Any help is appreciated.

Answer 1

Here are a couple possible approaches: pick your favorite.

• Show that $\bigoplus_{b \in B} S$ satisfies the universal property of $N \otimes_R S$, using the $R$-module homomorphism $N \to \bigoplus_{b \in B} S$, $\sum_{b \in B} \lambda_b b \mapsto (\lambda_b)_{b \in B}$.
• Use the universal property of $N \otimes_R S$ to conclude that for any $R$-linear functional $\phi : N \to R$, there exists a unique $S$-linear functional $\tilde\phi : N \otimes_R S \to S$ such that $\tilde\phi(x \otimes \lambda) = \lambda \cdot \phi(x)$ for $\lambda \in S$, $x \in N$. Now, if you have a basis $B$ for $N$, and $\sum_{b\in B} b \otimes \lambda_b = 0$, then apply this with the dual elements of $N^* = \operatorname{Hom}_R(N, R)$ to conclude that $\lambda_b = 0$ for all $b$.