Suppose $S$ is an $R$-algebra (associative, commutative, with unit...) such that $S$ is free of finite rank $n$ over $R$. Is it necessarily the case that we can find an $R$-basis $y\_1, ..., y\_n$ for $S$ with $y\_1 = 1$?


I can prove this when $n = 2$:


Suppose $x\_1, ..., x\_n$ is an $R$-basis for $S$, and write $1 = \sum\_{i=1}^n a\_ix\_i$, with $a\_i \in R$. We have $S$ faithfully flat over $R$, so we have $(a\_1, ..., a\_n) = (a\_1, ..., a\_n)S \cap R = (1)$, so we can find $b\_1, ..., b\_n$ in $R$ such that $1 = \sum\_{i=1}^n a\_ib\_i$. Now I use the assumption $n = 2$: the matrix $\left( \begin{array}{cc} a\_1 & a\_2 \\\ -b\_2 & b\_1 \end{array} \right)$ is unimodular, so $y\_1 = a\_1x\_1+a\_2x\_2,$ $y\_2 = -b\_2x\_1+b\_1x\_2$ is an $R$-basis of $S$ with $y\_1 = 1.$


More generally, as long as the unimodular row $(a\_1, ..., a\_n)$ can be completed to a unimodular matrix over $R$, we can use the unimodular matrix to produce an $R$-basis of $S$ with $y\_1 = 1$.


Since not every unimodular row can be completed to a unimodular matrix when $n \ge 3$ and $R$ is arbitrary, it seems like it may be possible to construct a counterexample, but I haven't been able to do it (specifying an algebra structure on a rank $3$ $R$-module is surprisingly tricky).



Does every finite free R-algebra have a basis starting with 1?
As Will Sawin says in his answer, if $R$ has a module $M$ such that the module $R\oplus M$ is free of rank 3 but $M$ is not free then there is a counterexample: $R\oplus M$ with zero multiplication in $M$.
For an example, let $R$ be the continuous (or just polynomial) functions on the $2$-sphere and let $M$ be the continuous (or polynomial) tangent vector fields.


Conversely, suppose $R$ has no such module. Whenever $S$ is an $R$-algebra such that $S$ is free of rank 3, then the ring map $R\to S$, considered as an $R$-module map, has a left inverse because the composed map $R\to S\to End\_R(S)$ has a left inverse. (There is an $R$-basis for the $3\times 3$ matrices over $R$ starting with the identity matrix.) Therefore as an $R$-module $S$ is the direct sum of $R$ and a free module; $S$ has a basis starting with $1$.


Note: We didn't need commutativity in $S$. We didn't need commutativity of $R$, either; we could just assume $R\to S$ is a homomorphism of associative unital rings for the last argument. But you did need commutativity of $R$ to say that if the module $M$ is such that $R\oplus M$ is free of rank $2$ then $M$ is free of rank one.


