# $m\mathbb{Z}\cdot \mathbb{Z}/n\mathbb{Z}$

I'm trying to find the tensor product $$\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}$$ using this theorem. Now I know what the end product is (namely: $$\mathbb{Z}/\gcd(m,n)\mathbb{Z}$$) and I proved it differently. However,

how do I get that $$m\mathbb{Z}\cdot \mathbb{Z}/n\mathbb{Z}\cong \gcd(m,n)\mathbb{Z}/n\mathbb{Z}$$?

• For my own learning, I am wondering what does $m\mathbb{Z} \cdot \mathbb{Z}/n\mathbb{Z}$ mean? It is not a tensor product?
– Mike
Commented Jul 6, 2019 at 1:54
• @Mike $\{k\cdot l \mid k\in m\mathbb{Z}, l\in \mathbb{Z}/n\mathbb{Z}\}$ Commented Jul 7, 2019 at 20:20

More generally, if $$I$$ and $$J$$ are ideals of the commutative ring $$R$$, we have $$I\cdot R/J=(I+J)/J$$ The proof is easier to formalize in the more abstract setting and is a simple verification.
In your case $$I=m\mathbb{Z}$$ and $$J=n\mathbb{Z}$$, so $$I+J=m\mathbb{Z}+n\mathbb{Z}=\gcd(m,n)\mathbb{Z}$$
• @UnexpectedExpectation Let $x\in I$ and $y\in R$; then $x(y+J)=xy+J\in (I+J)/J$. Let $x\in I$, $y\in J$. Then $(x+y)+J=x+J=x(1+J)\in I\cdot R/J$. Commented Jul 5, 2019 at 9:43