I'm having some troubles understanding a proof in Commutative Algebra Chapter I - VII of N. Bourbaki. It's on pag 114 of the book. Here's what it says:
Theorem 3
...
(ii) Conversely, if $M$ is an $A-$ projective module of rank 1, and $M^*$ is the dual of $M$, then the canonical homomorphism $u: M \otimes M^* \to A$ corresponding to the bilinear form $(x, x^*) \mapsto <x, x^*>$ on $M \times M^*$ is bijective.
Proof
It's sufficient to prove that, for every maximal ideal $\frak{m}$ of $A$, $u_\frak{m}$ is an isomorphism. As $M$ is finitely presented, $(M^*)_\frak{m}$ is canonically identified with $(M_\frak{m})^*$,and as $M_\frak{m}$ is free of rank 1 as its dual $(M_\frak{m})^*$ , clearly the canonical homomorphism $u_{\frak{m}}:(M_{\frak{m}}) \otimes (M_{\frak{m}})^* \to A_\frak{m}$ is bijective, which completes the proof.
What I don't really understand is the author seems to suggest that $u_\frak{m}$ is isomorphic, due to 2 facts: Firstly, $u_\frak{m}$ is canonical; and secondly, $(M_{\frak{m}}) \otimes (M_{\frak{m}})^*$ and $A_\frak{m}$ have the same rank (i.e, 1).
It's how I understand the paragraph, but it doesn't seem quite right to me. Of course, there maybe some homomorphisms $g: A_{\frak{m}} \to {A}_\frak{m}$ that aren't isomorphic (although both sides do have the same rank 1).
So I guess it must be because $u_\frak{m}$ is canonical. But I don't really see how a canonical homomorphism in this case must be an isomorphism? Can somebody please enlighten me. :'(
Or is there any other way to prove this?
Thank you very much in advance,
And have a good day, :x
$u_m$ is an isomorphism. To see this, if $M_m$ is free with basis, say $x$, then $M^*_m$ is free with dual basis $x^*$. Now check that $u_m(ax,bx^*)=ab$ is an isomorphism, here $x^*(x)=1$.