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I have the following problem:

Let $R$ be a ring, and take $g,g'$ in $R$ such that $g\neq g'$. Then $(g)=(g')$ iff there exists an invertible element $\beta \in R^\times$ such that $\beta g=g'$. (In our class we denoted by $(g)$ the ideal generated by $g$ and same for $g'$

So I wanted to prove this and my Idea was the following:

$\Leftarrow$ Let $\beta\in R^\times$ such that $\beta g=g'$. By definition $(g')=Rg=\{ag':a\in R\}$. Therefore $$(g')=(\beta g)=\{\beta a g:a\in R\}=\{bg:R\ni b=\beta a\}=(g)$$ Thus we have shown the first implication.

$\Rightarrow$ Let $(g)=(g') \Leftrightarrow \{ug:u\in R\}=\{vg':v\in R\}$. Then we know that $$g=vg'~~~\text{and}~~~g'=ug$$for some $u,v\in R$. But Why can I then say that $v$ is invertible or how do I find an invertible element $\beta$?

Can someone help me and maybe also correct the part I just did?

Thanks a lot.

user1294729
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1 Answers1

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We need to assume that $R$ is an integral domain. I am still looking for an example showing why this condition is required.

Your first implication is fine, thouh I would personally prefer a proof using double inclusion, among the following lines:

Assume $g=ug'$ for some unit $u$. To show $(g)\subseteq(g')$, let $ag\in(g)$; then $ag=uag'\in(g')$, as required. The other inclusion follows analogously, since $g'=vg$ for the unit $v=1/u$.

For the other direction, assume $(g)=(g')$. Then $g=ug'$ and $g'=vg$ for some $u,v$. Then $1\cdot g=uv\cdot g$, so $(1-uv)g=0$. As $R$ is integral, it holds that either $uv=1$, or $g=0$. This shows that either $u$ is a unit, or that $(g)=0$. The last case shows that also $g'=0$, hence $g=1\cdot g'$. In either case, $g$ is $g'$ times a unit.

Zuy
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