Short Exact Sequences and at Tensor Product

[WWGD]

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.

 

[micromass]

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.

 

[WWGD]

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.

 

[micromass]

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

$$\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}_2\cong \mathbb{Z}_2 \cong \mathbb{Z}_2 \otimes_\mathbb{Z}\mathbb{Z}_2$$

In general, we have the relation

$$\mathbb{Z}_n\otimes_\mathbb{Z} \mathbb{Z}_m \cong \mathbb{Z}_\text{gcd(m,n)}$$

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

 

[WWGD]

A, thanks, nice.