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Is it true that for a noncommutative $R$, a module $P$ is f.g projective iff $\mathsf{hom}(P,R)=P^\ast \cong P$?

Here's what I thought of as a proof: Since $(-)^\ast$ is additive, it preserves biproduct diagrams. The projective modules are the direct summands of free modules, and f.g projective modules can be taken as direct summands of f.g free modules. For finitely generated free modules the result is known. Then write $F=P\oplus Q$ for $F$ f.g free and apply the dualizing functor. Since its additive we have $P\oplus Q=F\cong F^\ast =P^\ast \oplus Q^\ast$ and this means (I think) that $P^\ast \cong P$ as desired. So it seems the only difference from the $P\cong P^{\ast\ast}$ is just that this isomorphism is not natural.

Is the above proof correct? What's the justification for inferring $P\cong P^\ast$ from the biproduct diagrams?

  • @QiaochuYuan I mean isomorphic to the dual. I think I'm missing something because it seems the only difference between the dual and the double dual is that for $1\cong (-)^{\ast\ast}$ everything is functorial, but for $P\cong P^\ast$ we still have isos, only pointwise. –  Nov 17 '15 at 23:40

1 Answers1

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If $R$ is noncommutative and $P$ is a right $R$-module then $\text{Hom}_R(P, R)$ naturally has the structure of a left $R$-module (and vice versa if $P$ is a left $R$-module), so you can't even ask for this isomorphism because the two objects belong to different categories.

But this is still false if $R$ is commutative. Take $R = \mathcal{O}_K$ to be the ring of integers of a number field $K$ whose ideal class group has a nontrivial element of order anything other than two. The corresponding ideal $I$ is f.g. projective and $I^{\ast}$ corresponds to the inverse of $I$ in the ideal class group; moreover, isomorphism of modules corresponds to equality of elements in the ideal class group.

Your proof fails because you don't ask for any compatibility between the isomorphism $F \cong P \oplus Q$ and the isomorphism $F \cong F^{\ast}$.

Qiaochu Yuan
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  • What do you mean by compatibility for the isomorphisms? –  Nov 17 '15 at 23:43
  • @Exterior: an isomorphism $F \cong F^{\ast}$ is the same thing as a nondegenerate bilinear form on $F$, and you should require that $P$ and $Q$ are orthogonal with respect to this bilinear form. The point is that this isn't always possible. Said another way, as written you get an isomorphism $P \oplus Q \cong P^{\ast} \oplus Q^{\ast}$ but there's nothing you can do to cancel terms from this because there's no guarantee that the two direct sum decompositions are compatible. And of course you need $R$ commutative in all of this. – Qiaochu Yuan Nov 17 '15 at 23:44
  • I don't quite understand what you're saying about the 'cancelling term' business. For instance, how is this not a problem for the double-dual isomorphism? There you have functorially $P\oplus Q\cong P^{\ast\ast}\oplus Q^{\ast\ast}$, but why does this resolve everything? –  Nov 17 '15 at 23:47
  • @Exterior: look, in general it's just not true that if $A \oplus B \cong A' \oplus B'$ then $A \cong A', B \cong B'$; you can find counterexamples for finite-dimensional vector spaces. It's not even true that if $B \cong B'$ then $A \cong A'$; you can find counterexamples using vector bundles which are stably trivial but not trivial. For the double dual isomorphism everything involved is functorial and so automatically respects biproducts. In this argument you've picked some arbitrary isomorphism $F \cong F^{\ast}$; there's no functoriality here and so no guarantee of nice behavior. – Qiaochu Yuan Nov 17 '15 at 23:49
  • Sorry for being slow! I understand functoriality is crucial now. Could you explain though why exactly we can deduce $P\cong P^{\ast\ast}$ from the biproduct isomorphism? –  Nov 17 '15 at 23:52
  • @Exterior: the double dual map $M \mapsto M^{\ast \ast}$ is a natural transformation between two functors, namely the identity functor and the double dual functor. This natural transformation is an iso for f.g. free modules, and both functors preserve biproducts, so the double dual map $P \oplus Q \to (P \oplus Q)^{\ast \ast} \cong P^{\ast \ast} \oplus Q^{\ast \ast}$ respects the biproduct structure on both sides. Now you just need to show that such a map between two biproducts is an iso iff its two components $P \to P^{\ast \ast}, Q \to Q^{\ast \ast}$ are isos. – Qiaochu Yuan Nov 18 '15 at 00:01
  • Thank you very much! Lastly, in the commutative case, is it at least true that $M\cong M^\ast$ iff $M$ is free and f.g, i.e can we at least generalize from vector spaces to modules over commutative rings? –  Nov 18 '15 at 00:03
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    @Exterior: no, that's not true either. In terms of my counterexample above, some ideal class groups do in fact have nontrivial elements of order $2$. There are also counterexamples coming from complex vector bundles: any complex vector bundle which is the complexification of a real vector bundle is self-dual, and most of these are not trivial. – Qiaochu Yuan Nov 18 '15 at 00:04
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    I see. I don't understand any of these examples, but I did ask about this in more detail in the second part of this question. –  Nov 18 '15 at 00:06
  • Last attempt - is it true that if $M$ is free f.g and $M^\ast$ is free then $M\cong M^\ast$? –  Nov 18 '15 at 09:54
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    @Exterior Of course: (if R is commutative,) $\hom_R(R^n,R)\cong R^n$, so the assumption that $M^\ast$ is free is redundant. The main problem you faced before is that $M\cong M^\ast$ simply doesn't imply freeness of M. – rschwieb Nov 18 '15 at 11:23