# Thread: Verification of tensor product

1. ## Verification of tensor product

Hi,

I just wondered if someone could verify my calculation that the tensor product of Z + Z/3 and Z/6 is isomorphic to Z/6 + Z/3.

The calculation is as follows:
(Z + Z/6) tensor Z/3 = (Z tensor Z/6) + (Z/3 tensor Z/6) = Z/6 + (Z/3 tensor Z/6)
Now, look at the exact sequence 0 -> Z -> Z -> Z/3 -> 0 with the homomorphisms f: Z -> Z, f(x)=3x and inclusion from Z to Z/3.
Tensor the sequence with Z/6. We get the exact sequence
0 -> Z/6 -> Z/6 -> Z/3 tensor Z/6 -> 0
so Z/3 tensor Z/6 is isomorphic to (Z/6)/f(Z/6) = (Z/6)/(Z/2) = Z/3, hence
Z/6 + (Z/3 tensor Z/6) = Z/6 + Z/3.

2. Originally Posted by claves
Hi,

I just wondered if someone could verify my calculation that the tensor product of Z + Z/3 and Z/6 is isomorphic to Z/6 + Z/3.

The calculation is as follows:
(Z + Z/6) tensor Z/3 ((Z + Z/3) tensor Z/6 ? ) = (Z tensor Z/6) + (Z/3 tensor Z/6) = Z/6 + (Z/3 tensor Z/6)
Now, look at the exact sequence 0 -> Z -> Z -> Z/3 -> 0 with the homomorphisms f: Z -> Z, f(x)=3x and inclusion from Z to Z/3.
Tensor the sequence with Z/6. We get the exact sequence
0 -> Z/6 -> Z/6 -> Z/3 tensor Z/6 -> 0
so Z/3 tensor Z/6 is isomorphic to (Z/6)/f(Z/6) = (Z/6)/(Z/2) = Z/3, hence
Z/6 + (Z/3 tensor Z/6) = Z/6 + Z/3.
If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is exact and G is torsion-free, then $0 \rightarrow A \otimes G \rightarrow B \otimes G \rightarrow C \otimes G \rightarrow 0$ is exact. In your case, the induced exact sequence is $\mathbb{Z}/6 \rightarrow \mathbb{Z}/6 \rightarrow \mathbb{Z}/3 \otimes \mathbb{Z}/6 \rightarrow 0$, because your original exact sequence is not a split short exact sequence.

We already know that $\mathbb{Z} \otimes G \cong G$. We can also use that $\mathbb{Z}/m \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n \mathbb{Z} \cong \mathbb{Z}/d \mathbb{Z}$ where d is the g.c.d of the integers m and n. To show this, you need to define a map $f:\mathbb{Z}/m \mathbb{Z} \times \mathbb{Z}/n \mathbb{Z} \rightarrow \mathbb{Z}/d \mathbb{Z}$ given by $f(\text{a mod m, b mod n)= ab mod d}$ is well-defined bilinear map. Then you need to show that the induced map $g:\mathbb{Z}/m \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n \mathbb{Z} \rightarrow \mathbb{Z}/d \mathbb{Z}$ sending $1 \otimes 1$ to $1 \in \mathbb{Z}/d \mathbb{Z}$ is an isomorphism. So we can directly compute $\mathbb{Z}/3 \mathbb{Z} \otimes \mathbb{Z}/6 \mathbb{Z} \cong \mathbb{Z}/3 \mathbb{Z}$.