3

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?

Bill Dubuque
  • 272,048
eChung00
  • 2,963
  • 8
  • 29
  • 42
  • It seems to be you're talking of ideals in the ring $;\Bbb Z;$ , so (1) Why not to mention that ? Why did you tag this question "group theory" instead of ring theory? – Timbuc May 20 '15 at 13:03
  • 1
    $m=nhk^{-1}$ is - when we are dealing with integers - no guarantee that $n$ divides $m$. However we have $mk\in(n)$ for any integer $k$. So just do it with $k=1$. On the second part: if $n$ divides $m$ then $m=nk$ for some $k$ (not $n=mk$). – drhab May 20 '15 at 13:06
  • @drhab yeah sorry. So if I let $m=nk$, then $(m) = { nkr\ |\ k,r \in \mathbb{Z} }$. So $(n)$ contains $(m)$. Is this good?? – eChung00 May 20 '15 at 13:13
  • Yes, that is correct. – drhab May 20 '15 at 13:14
  • 1
    You don't have to assume $R=\mathbb Z$. This holds for commutative rings too. – user222031 May 20 '15 at 13:19

2 Answers2

4

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 $\langle m\rangle\subset\langle n\rangle$.

Brian Fitzpatrick
  • 26,559
1

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.

Bill Dubuque
  • 272,048
  • What you mean with $R$? Please be more specific...(I guess that $R$ can be a general ring but needed to be commutative), and when you take $j=1=k$ we earn $a_1=c_{11}b_1$ ... – k1.M May 20 '15 at 17:10
  • @k1.M I presume $R,$ is a commutative ring. and obviously I use the abbreviation $, c = c_{11},$ to avoid ofuscatory notational clutter. Was that not clear? The downvote is ridiculous, but not surprising given the nonsensical downvoting nowadays. – Bill Dubuque May 20 '15 at 17:20
  • Now that's clear...but observe that you should write the assumptions in the answer not in the comments... – k1.M May 20 '15 at 17:23
  • 1
    @k1.M I expect the reader to have the mathematical maturity to make obvious sensible inferences (that's why it is a "hint"). – Bill Dubuque May 20 '15 at 17:24
  • what it means OP? – user441848 Jul 28 '17 at 21:44
  • @Annet. Original Post(er), either the initial post in a thread, or the author of it. – Bill Dubuque Jul 28 '17 at 21:44