Here is a simple fact: In your notation, the model under consideration is complete if and only if the matrix
1+R & 1+R \\
d & u
is one-to-one, i.e. $d \neq u$. (Equivalently, its transpose is onto, which is what is shown in your quoted text.)
No-arbitrage holds if and only if $d \leq 1+R \leq u$ and $d < u$. (By ...