Suppose $R\to S$ is a faithfully flat morphism of commutative unital rings, and suppose $A\to B\to C\to 0$ is an exact sequence of finitely-generated $R$-modules. If we know that $0\to A\otimes_R S\to B\otimes_R S\to C\otimes_R S\to 0$ is split exact, must it be the case that $A\to B\to C\to 0$ is also split exact?
I know that $0\to A\to B\to C\to 0$ must be exact by faithful flatness, but I am stumped on splitting. I feel like I ought to be able to come up with some inventive diagram or exact sequence which after tensoring would be exact due to splitting and conclude via faithful flatness, but defining this over $R$ seems problematic.
Best Answer
No, splitness does not descend. For example, consider $R \to S$ itself. Its base change is the $S$-linear map $S \to S \otimes_R S$ given by $s \mapsto 1 \otimes s$, which has an obvious retraction, namely $s_0 \otimes s_1 \mapsto s_0 s_1$. So, if $R \to S$ is faithfully flat, then $R \to S$ is monic. But there are examples where $R \to S$ is not a split monomorphism of $R$-modules.
For instance, take $R = \mathbb{Z}_{(6)} = \{ n / m \in \mathbb{Q} : n \in \mathbb{Z}, m \notin 6 \mathbb{Z} \}$ and $S = \mathbb{Z}_{(2)} \times \mathbb{Z}_{(3)}$. There is a unique ring homomorphism $R \to S$ and it is easily seen to be faithfully flat, but it does not split as a homomorphism of $R$-modules: to give a $R$-linear map $S \to R$ to split $S \to R$ is the same as giving $\mathbb{Z}_{(6)}$-linear maps $\mathbb{Z}_{(2)} \to \mathbb{Z}_{(6)}$ and $\mathbb{Z}_{(3)} \to \mathbb{Z}_{(6)}$ to split $\mathbb{Z}_{(6)} \to \mathbb{Z}_{(2)}$ and $\mathbb{Z}_{(6)} \to \mathbb{Z}_{(3)}$, respectively, but this is impossible because e.g. there is no element $x \in \mathbb{Z}_{(6)}$ such that $3 x = 1$ (unlike in $\mathbb{Z}_{(2)}$) or $2 x = 1$ (unlike in $\mathbb{Z}_{(3)}$).