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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!

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So you're the one who took the nym I originally wanted... –  anon Jul 23 '11 at 10:14
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@anon It's such an incredibly original nick, and it allows for such nice and personal communication, it's perfect for establishing a recognisable identity in the community... I can see why everybody would want to have it. –  Alex B. Jul 23 '11 at 13:46

1 Answer 1

up vote 5 down vote accepted

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 $R_s$.

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