Smooth ring maps and the module of differentials Suppose $A$ and $B$ are commutative Noetherian rings, and $A \to B$ is a finite type smooth map. Then it is well know that the module of Kähler differentials, $\Omega^1_{B/A}$ is a projective module of finite rank. It is also known that it is generated by elements of the form $db$ where $b \in B$.
My question: Is it true that locally there is a basis for the free module $\Omega^1_{B/A}$ of the form 
$\{db_1,\dots,db_n\}$?
Please correct me If I am wrong, but in general, a generating set for a free module does need need to have a subset which is a basis.
Thanks!
 A: 1) As you correctly conjecture, you cannot  in general extract a basis from a generating set of a free module. For example the set $\{2,3\}$ is a generating set for the free $\mathbb Z$-module $\mathbb Z $, and that set contains no basis.
2) Fact:  If a finitely generated projective $R$-module $P$ has generators $u_i (i\in I)$ , maybe in infinite number , then locally at any $\mathfrak p \in Spec(R)\;   $  a basis can be extracted.      
Proof: Wise use of Nakayama will  give an exact sequence of $R$-modules.
$$       0\to K\to  R^n \stackrel {u}{\to} P \to C\to 0       \quad (\ast)           $$
where $u$ is obtained from some of the given generators: $u(r_1,...,r_n)=\Sigma r_iu_i$ and where  $C_{\mathfrak p}=K_{\mathfrak p}=0$. [Wiseness means extracting generators $u_1,...,u_n$ that give a basis of the $\kappa (\mathfrak p)$- vector space$P\otimes_R \kappa (\mathfrak p)]$
Since the cokernel $C$ is a finitely generated $R$-module  it will be zero in some neighbourhood $D(r)$ of $\mathfrak p$. So we get the exact sequence on $D(r) \;$:
$$       0\to K_r\to  R_r^n \stackrel {u_r}{\to} P _r\to 0       \quad (\ast \ast)           $$
with $(K_r)_{\mathfrak p}=K_{\mathfrak p}=0$.
To apply the same trick to $K_r$ as we applied to  $C$ we must know that $K_r$ is finitely generated. But this is the case since $P$ is finitely generated projective, hence finitely presented. So finally we have locally on some open neighbourhoof $D(s)$ of $\mathfrak p$       
$$       0\to  R_s^n \stackrel {u_s}{\to} P _s\to 0       \quad (\ast \ast \ast)           $$
In other words $u_1,...,u_n$ is a basis of $P_s$ over $R_s$.
