1. ## Zm tensor Zn

This is a standard problem I've seen, but I am new to tensors. If I could see the proof I think it would help me greatly to compute tensors in general.

The problem statement here is from Hungerford GTM section IV.5.2

Show $Z_m \otimes Z_n \cong Z_c$ where c is gcd(m,n).

(everything is regarded as an add. abelian group-- the tensor is over Z)

A lead question is to show $A \otimes Z_m \cong A/mA$, which I think I managed with the map f sending $a \otimes (n+mZ) \mapsto (an+mA)$ showing the kernel was 0 and was surjective. For what it's worth, as a bijection this implies elements of $A \otimes Z_m$ are required to have the form simply $a \otimes 1$-- whether or not this will help with the problem stated above has not been something I can reconcile on my own.

Any insights would be appreciated greatly.

2. Originally Posted by gosualite
This is a standard problem I've seen, but I am new to tensors. If I could see the proof I think it would help me greatly to compute tensors in general.

The problem statement here is from Hungerford GTM section IV.5.2

Show $Z_m \otimes Z_n \cong Z_c$ where c is gcd(m,n).

(everything is regarded as an add. abelian group-- the tensor is over Z)

A lead question is to show $A \otimes Z_m \cong A/mA$, which I think I managed with the map f sending $a \otimes (n+mZ) \mapsto (an+mA)$ showing the kernel was 0 and was surjective. For what it's worth, as a bijection this implies elements of $A \otimes Z_m$ are required to have the form simply $a \otimes 1$-- whether or not this will help with the problem stated above has not been something I can reconcile on my own.

Any insights would be appreciated greatly.
To show $A \otimes Z_m \cong A/mA$, the first isomorphism theorem can be used.

First define a surgective map f such that

$f:A \otimes Z_m \rightarrow A$ given by $(a,x) \mapsto a$, where $a \in A , x \in Z_m$.

f is clearly surgective. Now, you need to find a kernel of f.

Since $ma \otimes x = a \otimes mx = a \otimes 0 = 0$, mA is a kernel of f.

To show $Z_m \otimes Z_n \cong Z_c$, you can use the above result and replace A with $Z_n$.

Since $Z_m \otimes Z_n \cong Z_n/mZ_n$, you remain to show that $Z_n/mZ_n \cong Z_c$. In a similar manner with the above, you can find a surgective map and its kernel and apply the first isomorphism theorem.

3. Originally Posted by aliceinwonderland
To show $A \otimes Z_m \cong A/mA$, the first isomorphism theorem can be used.

First define a surgective map f such that

$f:A \otimes Z_m \rightarrow A$ given by $(a,x) \mapsto a$, where $a \in A , x \in Z_m$.

f is clearly surgective. Now, you need to find a kernel of f.

Since $ma \otimes x = a \otimes mx = a \otimes 0 = 0$, mA is a kernel of f.
that's not correct! here's how to do it: the map $f_0: A \times \mathbb{Z}_m \longrightarrow A/mA$ defined by $f_0(a,k)=ka + mA$ is $\mathbb{Z}$ bilinear and thus it induces $f: A \otimes_{\mathbb{Z}} \mathbb{Z}_m \longrightarrow A/mA,$ which maps a generator

$a \otimes k$ to $ka + mA.$ now define $g:A/mA \longrightarrow A \otimes_{\mathbb{Z}} \mathbb{Z}_m$ by $g(a + mA)=a \otimes 1.$ it's clear that $g$ is a $\mathbb{Z}$ homomorphism. it's well-defined because if $a + mA=0,$ then $a=ma',$ and therefore

$a \otimes 1 = (ma') \otimes 1 = a' \otimes m=0.$ now we have $fg(a+mA)=f(a \otimes 1)=a+mA.$ hence $fg= \text{id}_{A/mA}$ and $gf(a \otimes k)=g(ka + mA)=(ka) \otimes 1=a \otimes k,$ i.e. $gf= \text{id}_{A \otimes \mathbb{Z}}. \ \Box$

... you remain to show that $Z_n/mZ_n \cong Z_c$.
we have: $m(\mathbb{Z}/n\mathbb{Z})=(m\mathbb{Z}+n\mathbb{Z})/n \mathbb{Z}=c \mathbb{Z}/n \mathbb{Z}.$ thus $\mathbb{Z}_n / m \mathbb{Z}_n = (\mathbb{Z}/n \mathbb{Z})/(c \mathbb{Z}/n \mathbb{Z}) \cong \mathbb{Z}/c \mathbb{Z}=\mathbb{Z}_c.$

Notation: you should use $\mathbb{Z}/n$ or $\mathbb{Z}/n \mathbb{Z}$ instead of $\mathbb{Z}_n.$

Remark: a trivial generalization of this problem is that for any unitary commutative ring $R,$ ideals $I,J$ and $R$ module $A$ we have: $\frac{R}{I} \otimes_R A \cong \frac{A}{IA},$ and $\frac{R}{I} \otimes_R \frac{R}{J} \cong \frac{R}{I+J}.$

by the way i'm not completely back yet! i'll try to help whenever i can (unfortunately not as often as before! don't blame me, blame my advisor!)

we have: $m(\mathbb{Z}/n\mathbb{Z})=(m\mathbb{Z}+n\mathbb{Z})/n \mathbb{Z}=c \mathbb{Z}/n \mathbb{Z}.$