Find $n\mathbb{Z}/m\mathbb{Z}$ where $n|m$

Question

I have the following example in the lecture notes I'm reading: $n\mathbb{Z}/m\mathbb{Z}$ where $n|m$, where the abelian groups $n\mathbb{Z},m\mathbb{Z}$ are examples of $\mathbb{Z}$-modules. It's stated that $n\mathbb{Z}_{m}=n(\mathbb{Z}/m\mathbb{Z})=n\mathbb{Z}/m\mathbb{Z}$. However I don't understand why $n(\mathbb{Z}/m\mathbb{Z})=n\mathbb{Z}/m\mathbb{Z}$. Doesn't the first factor module represent $n\mathbb{Z}/\frac{m}{n}\mathbb{Z}$?

Answer 1

Well one way (the straight-forward using definition one) to look at it is that:

an element of $n(\mathbb{Z}/m\mathbb{Z})$ is of the form $n(k+m\mathbb{Z})$ where $k\in \mathbb{Z}$.

But $n(k+m\mathbb{Z})=nk+m\mathbb{Z}$ by the addition operation of the quotient ring $\mathbb{Z}/m\mathbb{Z}$.

Finally, $nk+m\mathbb{Z}\in n\mathbb{Z}/m\mathbb{Z}$.

Answer 2

take the map $\phi_n:\mathbb Z \to \mathbb Z$ given by multiplication by $n$, and take the composition with the quotient map $\rho:\mathbb Z \to \mathbb Z_m$, so that $\mathrm{Im}(\rho \circ \phi_n)=n\mathbb Z/\mathbb Z_m$. But by the universal property of the quotient map, there is an induced map $\mathbb Z_m \to \mathbb Z_m$ given by multiplication by $n$. The image of this map is $n(\mathbb Z_m)=n (\mathbb Z/m\mathbb Z)$.

Answer 3

The notation is perhaps confusing. Let's look at a more general setting: we have a commutative ring $A$ and an ideal $I$ of $A$.

If $a\in A$, then $$ a(A/I)=(aA+I)/I $$ because the elements of $A/I$ are of the form $x+I$. If also $I\subseteq aA$, then $a(A/I)=aA/I$.

Now translate it for $A=\mathbb{Z}$, $a=n$ and $I=m\mathbb{Z}$. Since $n\mid m$, we have indeed $m\mathbb{Z}\subseteq n\mathbb{Z}$.

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The general case, since $n\mathbb{Z}+m\mathbb{Z}=k\mathbb{Z}$, where $k=\gcd(n,m)$, is then $$ n(\mathbb{Z}/m\mathbb{Z})=(n\mathbb{Z}+m\mathbb{Z})/m\mathbb{Z}= k\mathbb{Z}/m\mathbb{Z} $$ (and in case $n\mid m$, $k=\gcd(m,n)=n$).

By the way, this group is isomorphic to $\mathbb{Z}/(m/k)\mathbb{Z}$, so in general $$ n(\mathbb{Z}/m\mathbb{Z})\cong\mathbb{Z}/m'\mathbb{Z} $$ where $$ m'=\frac{m}{\gcd(n,m)} $$