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Let $R$ be a ring and $r,s∈R$

So this is true

If $r=st$ for some $t∈R^x$, then $(r)=(s).$

$R^x$ being the units in $R$

Why is it true and why is the converse not?

If $(r)=(s)$, then $r=st$ for some $t∈R^x$

My thought process for part 1 is $r = st$ then $r ∈ (s)$ by the definition of an ideal. Then we can just generate the rest of $(r)$ so $(r) ⊆ (s)$. Also if t is a unit the $rt^{-1} = s$ and by the same logic $(s) ⊆ (r)$

Bing Bong
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  • Your argument about the first part is correct. I'll see if I can come up with a simple counter-example to the second part. – Rob Arthan May 05 '22 at 18:19
  • Sorry, could you explain it a bit I'm confused about gc in the poster's example does not imply 1 I thought a unit by definition has the multiplicative inverse property? Obviously, this doesn't work if the ring doesn't have units but would this not work for every ring with units? – Bing Bong May 05 '22 at 18:36
  • @BingBong Not sure what you are talking about. The post at least links to duplicates that give examples of rings in which $(r)=(s)$ but $r\neq st$ for any unit $t$. For example this one: https://math.stackexchange.com/a/3019359/29335 and https://math.stackexchange.com/q/14270/29335 – rschwieb May 05 '22 at 19:17

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