# Thread: tensor product

1. ## tensor product

If G is finite Abelian group and G (X)_Z Z/pZ={0} ( tensor of G with Z/pZ over Z )
for all primes p, then show that G = {0}. Does the result remain true if G is infinite ?

Since G is finite abelian then G= Z/n_1Z x Z/n_2Z x ... x Z/n_kZ
where n_k | n_k-1 | ... | n_2 | n_1
so {0}= G(X)Z/pZ = Z/dZ where d=gcd(n_k , p) for all primes
then d=1 and for each p there is one of n_i relatively prime to p
but can this imply G is zero ?
Also, I have no idea about the second assertion.

2. Originally Posted by hgd7833
If G is finite Abelian group and G (X)_Z Z/pZ={0} ( tensor of G with Z/pZ over Z )
for all primes p, then show that G = {0}. Does the result remain true if G is infinite ?

Since G is finite abelian then G= Z/n_1Z x Z/n_2Z x ... x Z/n_kZ
where n_k | n_k-1 | ... | n_2 | n_1
so {0}= G(X)Z/pZ = Z/dZ where d=gcd(n_k , p) for all primes
then d=1 and for each p there is one of n_i relatively prime to p
but can this imply G is zero ?
Also, I have no idea about the second assertion.

The second assertion is easy to show is false: take $\mathbb{Q}\otimes_\mathbb{Z} G$ , with $G$ any finite abelian group, say of order

$n\Longrightarrow q\otimes g=n(q/n\otimes g)=q/n\otimes ng=q/n\otimes 0=0$ ...

Tonio

3. No. The question is : what if G is infinite , you're assuming G is finite in your argument.

4. Originally Posted by hgd7833
No. The question is : what if G is infinite , you're assuming G is finite in your argument.

No, $\mathbb{Q}$ is infinite and Z-tensored with any finite abelian group, and in particular with $\mathbb{Z}/p\mathbb{Z}$ we get the zero group, thus showing that an infinite group can give zero tensored with any group of order a prime p and nevertheless it is not, obviously, the zero group.

Tonio

5. Oh yes, so Q is your group. This works now.
Did you look at my argument for the first part ? Is it correct ? Thanks

6. for the first part, suppose to the contrary that $G \neq \{0\}.$ then $|G| \geq 2$ and $\{0\}=G \otimes \mathbb{Z}/p \mathbb{Z} \cong \bigoplus_{i=1}^k (\mathbb{Z}/n_i \mathbb{Z} \otimes \mathbb{Z}/p \mathbb{Z}).$ thus $\mathbb{Z}/n_i \mathbb{Z} \otimes \mathbb{Z}/p \mathbb{Z}=\{0\},$ for all $i$ and all primes $p.$

in particular $\mathbb{Z}/n_1 \mathbb{Z} \otimes \mathbb{Z}/p \mathbb{Z}=\{0\}$ for all primes $p.$ but that is false because if $p \mid n_1,$ then $\mathbb{Z}/n_1 \mathbb{Z} \otimes \mathbb{Z}/p \mathbb{Z} \cong \mathbb{Z}/p\mathbb{Z} \neq \{0\}.$

7. Yes I got it now. Thanks very much to you and to Tonio !