# Inversion of an element in Picard group over commutative ring

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 $$(x, x^*) \mapsto $$ 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 $x^*(x)=1$.
• Indeed, the point is that localization turns projective modules into free modules, so you can use a basis and the dual basis to write the image of $u_m$ explicitly. Nov 10, 2014 at 5:51
• Great, thank you very much, I think I get it. It's easy to see that $u_\frak{m}$ is epic, since $u_{\frak{m}}(ax; x^*) = a, \forall a \in A_{\frak{m}}$. And it's also mono, say $0 = u_{\frak{m}}\left( \sum a_ix \otimes b_ix^* \right) = \sum a_ib_i$, so, we'll have $\sum a_ix \otimes b_ix^* = \left[ \sum \left( a_ib_i\right) \right] x \otimes x^* = 0 \otimes x^* = 0$. Does this look correct? Thank you very much :D Nov 10, 2014 at 7:55