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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

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(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

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1 Answer 1

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$u_m$ 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 $x^*(x)=1$.

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  • $\begingroup$ 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. $\endgroup$
    – guest
    Nov 10, 2014 at 5:51
  • $\begingroup$ 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 $\endgroup$
    – user49685
    Nov 10, 2014 at 7:55

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