prove $(n) \supseteq (m)\iff n\mid m\ $ (contains = divides for principal ideals)

Question

For non-zero integers $m$ and $n$, prove $(m) \subset (n)$ iif $n$ divides $m$, where $(n)$ is the principal ideal. My attempt is following.

For non-zero integers $m$ and $n$, assume that $(m) \subset (n)$. Then, $mk \in (m)$ is also in $(n)$. This means that $\exists nh$ such that $mk = nh.$ Then, we have $m=nhk^{-1}.$

Assume that $n$ divides $m$ for non-zero $n$ and $m$. Then, $n = mk$ and the principal ideal of $n$ will be $(n)=\{ mkr\ |\ n,r \in \mathbb{Z} \}.$ Since $kr \in \mathbb{Z}$, clearly $(m) \subset (n)$. is this ok proof?

Answer 1

Let $m,n\in\Bbb Z$. We wish to show that $\langle m\rangle\subset\langle n\rangle$ if and only if $n\mid m$.

First, suppose that $\langle m\rangle\subset\langle n\rangle$. Since $m\in\langle m\rangle$ it follows that $m\in\langle n\rangle$. That is, there exists a $k\in\Bbb Z$ such that $nk=m$. Hence $n\mid m$.

Conversely, suppose that $n\mid m$. To show that $\langle m\rangle\subset\langle n\rangle$, let $km\in\langle m\rangle$. Then, since $n\mid m$, there exists an $\ell\in\Bbb Z$ such that $\ell n=m$. It follows that $km=k\ell n\in\langle n\rangle$. Hence $\langle m\rangle\subset\langle n\rangle$.

Answer 2

Hint $\,\ (a_1,\ldots,a_j) \subseteq (b_1,\ldots,b_k)\!\!\overset{\rm\color{#c00}{U}\!\!}\iff a_1,\ldots,a_j \in (b_1,\ldots,b_k)\iff $

$$\iff \begin{bmatrix} a_1\\ \vdots\\ a_j\end{bmatrix}\,=\,\begin{bmatrix}c_{11} &\ldots &c_{1k}\\ \vdots &\ddots & \vdots\\ c_{j1}&\ldots &c_{jk} \end{bmatrix}\begin{bmatrix} b_1\\ \vdots\\ b_k\end{bmatrix}\ \ \text{for some }\ c_{i,j}\in R$$

OP is case $\, j = 1 = k,\ $ i.e. $\ a_1 = c\, b_1\ $ for some $\ c\in R,\ $ i.e. $\ b_1\!\mid a_1,\, $ i.e.

$$\quad\ (b_1)\supseteq (a_1) \iff b_1\mid a_1\qquad {\bf [contains = divides]}$$

Here $\rm\color{#c00}{U} =$ universal property of ideal sum: $\, A_1+A_2 \subseteq B\!\iff\! A_1,A_2\subseteq B,\,$ for any ideal $\,B,\,$ i.e. the ideal sum is the smallest ideal containing the summands. Above is the special case when the summands $A_i = (a_i)$ are principal ideals.

$\!\!\begin{align}\text{For proper containment:}\ \ (a)\supsetneq (b)&\iff (a)\supseteq (b)\text{ and not }(b)\supseteq (a)\\[.4em] &\iff\ \ a\ \ \mid\ \ b\ \ \text{ and not }\ \ b\ \ \mid\ \ a \end{align}$

which means that $\,(a)\supset (b)\,$ properly $\iff\, a\mid b\,$ properly, i.e. the "proper" analog holds.