I have an integral domain $R$ with field of fractions $K$. Let $M$ be a finitely generated free module over $R.$ If $M$ has an $R$-basis $\{u_i\}_{i=1}^d$ then is it true that the set $\{1\otimes_R u_i\}_{i=1}^d$ is a $K$-basis of the vector space $V:=K\otimes_RM$?
I can see that the set spans $V$ over $K$ but I'm not sure why it's linearly independent over $K$.
If $K$ is the field of fractions of $R$, or Körper as we say in german, then we have $K\otimes_R R=K$. Now since $M$ is free, we have $M\simeq \bigoplus_{i=1}^n R$ this means that
$$K\otimes M\simeq \bigoplus_{i=1}^n K\otimes_R R\simeq \bigoplus_{i=1}^n K$$