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Let $I$ be an ideal of $\mathbb{Z}$. Prove that $I = (m)$ for some $m ∈ \mathbb{Z}$.

My Attempt:

So $(m)=\{ am \, | \, a \in \mathbb{Z} \}$

If $m=0$ then $I=\{ 0 \}$ which is trivial.

1) If $m>0$, let $A \in I$ be the maximum in $I$. By division theorem, $A=xm+r$ where $x,r \in \mathbb{Z}$ and $0 \leq r <m$. If $r$ is anything besides zero, then $A$ wouldn't be the maximum. So $r=0$ by assumption, so $A=xm \in (m)$. So $I$ is a subset of $(m)$.

2) Let $am, bm \in (m)$ where $a,b$ are integers, then $am-bm= (a-b)m \in I$ since $a-b \in \mathbb{Z}$ and $I$ is an ideal of $\mathbb{Z}$. So $(m)$ is a subset of $I$.

3) If $m<0$, let $m=-n$ for some $n \in \mathbb{N}$ and let $A \in I$ be the minimum in $I$. By division theorem, $A=xm+r=-xn +r$ where $x,r \in \mathbb{Z}$ and $-n < x < \leq $. If $r$ is anything besides zero, then $A$ wouldn't be the minimum. So $r=0$ by assumption, so $A=-xn=xm \in (m)$. So $I$ is a subset of $(m)$.

4) Same as (2). So $(m)$ is a subset of $I$.

Hence $I=(m)$.

Now I know that for part 1 and 3, I was meant to say min for 1 and max for 3 which was a silly mistake I did in my exam. Also in 1 and 3, I didn't rearrange to $r$ so I could imagine this as lack of justification.

But please would you estimate a mark out of 7 for this. Surely I at least got one mark...

snowman
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1 Answers1

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Let $I$ be an ideal of $\mathbb Z$. If $I$ is trivial, then $I=(0)$. Otherwise let $k\in I\setminus\{0\}$. If $k<0$, then $0<-1\cdot k\in I$, so $I$ has a positive element. Let $m$ be the least positive element of $I$ and note that $(m)\subseteq I$ by the ''absorption property'' of ideals.

Assume to the contrary that there is an $n\in I$ such that $n\neq a\cdot m$ for all $a\in\mathbb Z$. So $n=mq+r$ for some $0<r<m$ and $q\in\mathbb Z$ by the quotient-remainder theorem. Now $q\cdot m\in I$ and $I$ is a subring of $\mathbb Z$, so $n-q\cdot m=r\in I$. But $0<r<m$ contradicts our assumption that $m$ was the least positive element of $I$. Thus no such $n$ can exist, and $I\subseteq(m)$.

Hence $I=(m)$ as desired.

User12345
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  • hi, just had my exam and this question came up but m was an integer - not a natural number. may I just ask, would it be OK for me to post what I did for it on this thread and you (or anyone else) mark it out of 7 marks please? I would really appreciate it. I wanted to ask you first before I post it so I wouldn't waste time. – snowman May 14 '15 at 21:23
  • The stack exchange encourages users to post answers to their own questions. I'll tell you what I think of your proof if you'd like, but I would rather not assign it a grade. – User12345 May 14 '15 at 21:53
  • It is up on the main post. I would really appreciate it if you gave a mark out of 7 though because I cant stop thinking about this question!!! – snowman May 14 '15 at 22:19