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

Question

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, Jul 06 '19 at 01:54
@Mike ${k\cdot l \mid k\in m\mathbb{Z}, l\in \mathbb{Z}/n\mathbb{Z}}$ — bliipbluup, Jul 07 '19 at 20:20

score 3 · Accepted Answer · answered Jul 05 '19 at 08:55

3

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 $$ I+J=m\mathbb{Z}+n\mathbb{Z}=\gcd(m,n)\mathbb{Z} $$

answered Jul 05 '19 at 08:55

egreg

244,946

Where can I find a proof for this general case? – bliipbluup Jul 05 '19 at 09:13
1

@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 $(x+y)+J=x+J=x(1+J)\in I\cdot R/J$. – egreg Jul 05 '19 at 09:43

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

1 Answers1