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While reading Dieudonné's Treatise on Analysis III", I studied tensor products only from his annexes A10 and A11 on algebra in this same volume. This is not very exhaustive. But I need only some results from algebra in order to understand his proofs in analysis. This is difficult enough for me, so presently I have not the time do dive deeply into multilinear algebra. So here are my questions:

Let A be any commutative ring with 1, and let E, F be A-modules, finite dimensional.

There is a well-known isomorphism

$ E^{\star} \otimes F \cong Hom(E,F) \qquad $ (1)

which Dieudonné explains and deduces in A.10.5. There is also a good proof here. But Dieudonné in 16.5.8.3 also uses this one

$ F \otimes E^{\star} \cong Hom(E,F) \qquad $ (2)

which he does not explain in the annex. Is it true? I just want to be sure, because consequently we also have the isomorphism

$ E^{\star} \otimes F \cong F \otimes E^{\star} \qquad $ (3)

which Dieudonné seems to have implied in A11 but never mentions explicitly. Now what about this one, which to me seems to be deducible in the analogous way as Dieudonné deduces (1), that is, by letting $ E_1 = F_2 = A $ in A.10.5.3:

$ E \otimes F^{\star} \cong Hom(E,F) \qquad $ (4).

Is it true? Actually, I doubt it, because it would imply

$ E^{\star} \otimes F \cong E \otimes F^{\star} \qquad $ (5).

And finally, is there a natural isomorphism

$ Hom(A,E) \cong E \qquad $ (6)?

My approach is with (1)

$ Hom(A,E) \cong A^{\star} \otimes E \cong A \otimes E \cong E \qquad $ (7).

From the information in Dieudonnés annex I don't find the answers. Thanks for your help to someone who is not studying math.

Roland Salz
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1 Answers1

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I guess $A$ is a commutative unital algebra over a field.
Then the tensor product is commutative as well (up to isomorphism): $M\otimes N\cong N\otimes M$ via the obvious map $m\otimes n\mapsto n\otimes m$.

So, (2) and (3) hold.
However, (4) and (5) do not.

To answer your final question, your proof is fine, however we can directly define an isomorphism $\hom_A(A,E)\to E$ by $f\mapsto f(1)$.

Berci
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  • Sorry for having been imprecise: A is any commutative ring with 1. – Roland Salz Oct 21 '20 at 17:48
  • Yes, it's working for that case too. – Berci Oct 21 '20 at 20:15
  • Thanks a lot for your answer. The first part is clear to me. What I don't understand in your last part: How do we know that your map $ f \mapsto f(1) $ is bijective? What is the inverse mapping? – Roland Salz Oct 21 '20 at 21:39
  • It's $e\mapsto,(a\mapsto ae)$. The point is that the $A$-module morphisms from $A$ are already determined by the image of the identity element. (Putting it other way, $A$ is the free module on one generator.) – Berci Oct 21 '20 at 22:01
  • I think I understand. Thanks again for your kind help! – Roland Salz Oct 22 '20 at 12:02