Timeline for Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$?

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Dec 5, 2015 at 8:29	comment	added	Owen Biesel		Thanks! At some point I started thinking of finite projective algebras as not just locally free, but locally admitting a basis that contains 1. The proof I eventually came across was that since $R\hookrightarrow S$ is a pure submodule and $S$ is flat, the quotient $S/R$ is also flat. It's also finitely presented if $S$ is, hence projective, hence the exact sequence $0\to R\to S\to S/R \to 0$ splits.
Dec 4, 2015 at 17:59	history	answered	John Voight	CC BY-SA 3.0