Short Exact Sequences and at Tensor Product

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Discussion Overview

The discussion centers around the properties of short exact sequences of Z-modules and their behavior under the tensor product with a ring R. Participants explore whether the tensor products of these modules remain isomorphic when the original modules are isomorphic, particularly in the presence of torsion.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that for a short exact sequence of Z-modules, if A and B are isomorphic, then their tensor products A(x)R and B(x)R are also isomorphic, but questions this in the context of torsion in R.
  • Another participant asserts that A(x)R and B(x)R are indeed isomorphic, inviting others to seek a proof for this claim.
  • A later reply clarifies that the initial inquiry was misunderstood, and instead seeks an example where A(x)R and B(x)R are isomorphic while A and B are not isomorphic.
  • One participant provides an example involving tensor products of Z-modules, specifically mentioning the relation between tensor products and the greatest common divisor, suggesting this could yield counterexamples.

Areas of Agreement / Disagreement

Participants express differing views on the isomorphism of tensor products in relation to the original modules, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

The discussion involves assumptions about the properties of modules and tensor products, particularly in the context of torsion and exact sequences, which may not be fully explored or agreed upon.

Who May Find This Useful

Readers interested in module theory, exact sequences, and the properties of tensor products in algebra may find this discussion relevant.

WWGD
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Hi,let:

0->A-> B -> 0

; A,B Z-modules, be a short exact sequence. It follows A is isomorphic with B.

. We have that tensor product is

right-exact , so that, for a ring R:

0-> A(x)R-> B(x)R ->0

is also exact. STILL: are A(x)R , B(x)R isomorphic?

I suspect no, if R has torsion. Anyone have an example of

A(x)R , B(x)R non-isomorphic, but A,B isomorphic? Thanks.
 
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WWGD said:
Hi,let:

0->A-> B -> 0

; A,B Z-modules, be a short exact sequence. It follows A is isomorphic with B.

. We have that tensor product is

right-exact , so that, for a ring R:

0-> A(x)R-> B(x)R ->0

is also exact. STILL: are A(x)R , B(x)R isomorphic?

Yes, they are. You should look for a proof.
 
Sorry, that is not what I meant to ask, instead , I am looking for an example where:

A(x)R , B(x)R are isomorphic,

but A,B are not isomorphic.

Thanks.
 
WWGD said:
Sorry, that is not what I meant to ask, instead , I am looking for an example where:

A(x)R , B(x)R are isomorphic,

but A,B are not isomorphic.

Thanks.

Oops, I never saw your reply. Weird. Anyway, consider

[tex]\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}_2\cong \mathbb{Z}_2 \cong \mathbb{Z}_2 \otimes_\mathbb{Z}\mathbb{Z}_2[/tex]

In general, we have the relation

[tex]\mathbb{Z}_n\otimes_\mathbb{Z} \mathbb{Z}_m \cong \mathbb{Z}_\text{gcd(m,n)}[/tex]

This result will also give you a wealth of counterexamples (note that the above still holds if we define ##\mathbb{Z}_0 = \mathbb{Z}##)
 
A, thanks, nice.
 

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