Locally free module

Question

I've got an $A$-module $M$, where $A$ is commutative. I should prove that $M$ is a locally free module of rank one iff $M^* \otimes M\cong A$ via the standard trace application. (For me locally free of rank one means that there is a covering of SpecA by$X_f$ such that $M_f \cong A_f$ for every such $f$) One arrow is clear to me: if the module is locally free of rank one,one studies the localized applications and here it is not difficult to prove. The other arrow is really an enigma to me: I do not have any idea how to proceed.

Answer 1

Let $f:M^*\otimes M\stackrel {\cong}{\to} A:\mu\otimes m\mapsto \mu(m)$ be the canonical morphism and assume it is an isomorphism.
So there exist two families $\mu_i\in M^*, m_i\in M (i=1,\cdots,r)$ such that $$f(\sum\mu_i\otimes m_i)=\sum\mu_i( m_i)=1\in A$$ Consider now the morphism $A^r\to M:e_i\mapsto m_i$ and tensor it with the identity of $M^*$, thus obtaining a morphism $A^r\otimes M^*\to M\otimes M^*:e_i\otimes \mu\mapsto m_i\otimes \mu$ which you can compose with the isomorphisms $ M\otimes M^*\cong M^*\otimes M\cong A$, t0 obtain the morphism$$g:A^r\otimes M^*\to A:e_i\otimes \mu\mapsto \mu(m_i)$$ That morphism $g$ is surjective since $g (\sum e_i\otimes \mu_i)=\sum\mu_i( m_i)=1$.
Thus (since $A$ is $A$-projective !) we can write $A^r\otimes M^*=A\oplus J\;$ ($J$ for Junk!).
Tensoring with $M$ on the right, we get $A^r\otimes M^*\otimes M=M\oplus (J\otimes M)$.
Hence (using the isomorphism $f$ again) we get $$A^r=M\oplus (J\otimes M)$$ This proves that $M$ is finitely generated and projective of rank one [rank is computed by localizing at the primes of $A$] , which is well known to imply that $M$ is locally free of rank one.