Basis of a finite free $R$-algebra starting with $1$ , $R$ is local Let be $A$ be a finite free $R$-algebra where $R$ is a local ring. Does $A$ always have a basis $a_1, \ldots, a_n$, where $a_1=1$?

Yes, that's true: More generally, if $k:=R/{\mathfrak m}$ is the residue field of $R$ and if $M$ is a finite free $R$-module such that $\{\overline{m_1},\ldots,\overline{m_n}\}$ is a basis of $M\otimes_R k=M/{\mathfrak m}M$, then ${\mathscr B}:=\{m_1,\ldots,m_n\}$ is a basis of $M$. For the proof, note that the base change of ${\mathscr B}$ to any fixed basis ${\mathscr C}$ of $M$ has non-vanishing determinant in $k=R/{\mathfrak m}$, hence is invertible in $R$ since $R$ is local. 
