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Timeline for Tensor product and homomorphism

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Feb 7, 2023 at 23:04 history became hot network question
Feb 7, 2023 at 22:04 vote accept marco2013
Feb 7, 2023 at 20:53 answer added Martin Brandenburg timeline score: 8
Feb 7, 2023 at 20:50 comment added Benjamin Steinberg @MartinBrandenburg, maybe it isn't. I'm used to finite dimensional algebras
Feb 7, 2023 at 20:47 comment added Martin Brandenburg I also highly doubt that this is sufficient.
Feb 7, 2023 at 20:45 comment added Benjamin Steinberg Maybe one should ask M to be finitely presented to be safe
Feb 7, 2023 at 20:36 comment added Martin Brandenburg @BenjaminSteinberg Maybe I misunderstood your comment, but you are saying that $f$ is always injective? I doubt that.
Feb 7, 2023 at 20:07 comment added Benjamin Steinberg I think it is injective and the image is all endomorphisms M that factor through a finitely generated projective. I don't have time to check details but the inverse map should involve choosing a factorization through some $A^m$ and then use the components to build something in the tensor product. One would need to show well defined
Feb 7, 2023 at 17:19 history edited marco2013
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Feb 7, 2023 at 16:11 comment added marco2013 Yes, if $M$ is projective, $f$ is a monomorphism, but the reverse is not true. For example, if $A=\mathbb{Z}$ and $M=\mathbb{Z}/n \mathbb{Z}$, then $M^*=0$, so $M \otimes M^*=0$, so $f$ is injective, but $M$ is not a projective module.
Feb 7, 2023 at 15:50 comment added Sampah It is a classical result that the mentioned map is a bijection if and only if $M$ is finitely generated and projective (fgp), so being fgp is a sufficient condition. But I believe we can weaken it to the module $M$ being merely projective.
Feb 7, 2023 at 15:41 review Close votes
Feb 7, 2023 at 21:57
Feb 7, 2023 at 15:04 history asked marco2013 CC BY-SA 4.0