# Thread: Proof: Z/nZ (x) Q = 0

1. ## Proof: Z/nZ (x) Q = 0

Hello,
I have to proof that: $\mathbb Z/n\mathbb Z \otimes \mathbb Q = 0$
I proofed it as follows but I don't know if it is correct:

${\text{Let }}a \in \mathbb{Z},{\text{ }}b \in \mathbb{Q}.$

$(a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\$
$= (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}})
= abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\$

$= (0 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \cdot (1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \hfill \\$

Please correct me, if there are any mistakes.
Thank you!

-Lisa-

2. Its correct (if it is $\otimes_{\mathbb{Q}}$).
If it is $\otimes_{\mathbb{Z}}$, then you should rewrite $b$ as a fraction of integers, since you factor the fraction $b$ out of the tensor product. But that is an easy modification.

3. Originally Posted by lisa
Hello,
I have to proof that: $\mathbb Z/n\mathbb Z \otimes \mathbb Q = 0$
I proofed it as follows but I don't know if it is correct:

${\text{Let }}a \in \mathbb{Z},{\text{ }}b \in \mathbb{Q}.$

$(a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\$
$= (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}})
= abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\$

$= (0 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \cdot (1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \hfill \\$

Please correct me, if there are any mistakes.
Thank you!

-Lisa-
it's correct. you can also write: $(a + n \mathbb{Z}) \otimes r = (a + n\mathbb{Z}) \otimes (n \cdot n^{-1}r)=(na + n\mathbb{Z}) \otimes n^{-1}r = 0 \otimes n^{-1}r = 0.$

4. Hi,

thank you for your answer.

5. Originally Posted by vemrygh

Its correct (if it is $\otimes_{\mathbb{Q}}$).
it has to be $\otimes_{\mathbb{Z}}$ because $\mathbb{Z}/n\mathbb{Z}$ is not a $\mathbb{Q}$ module.