0

I proved that in an integral domain that the implication always holds, but apparently it doesn't necessarily hold in all rings, and I have been looking for counterexamples. But, I'm confused why this proof that doesn't use the assumption of $R$ being an integral domain fails (and certainly it fails since I'm "proving" a false result):

Let $R$ be a ring, not necessarily an integral domain. Say $$ (a) = (b) $$ Thus, $a\in (b)$ and therefore $a = b\circ c$ for some $c\in R$.

Further, $b\in (a)$ and similarly $b = a\circ g$.

Thus, plugging in $b = a\circ g$, we get $a = a\circ g\circ c$. Hence, $g\circ c = 1$, so $g, c\in R^*$. Thus, letting $u = c$, we get $a = b\circ u$ for some unit $u$.

I'm unsure why this proof fails.

user26857
  • 52,094
  • $a=agc$ does not imply $gc=1$. This is true in integral domains, but can fail in general. – user26857 Nov 29 '21 at 05:55
  • In $\mathbb Z/6\mathbb Z,$ $\langle 2\rangle=\langle 4\rangle.$ $g=c=2.$ – Thomas Andrews Nov 29 '21 at 05:59
  • (Nonzero) rings that are not domains always have nonzero elements that are not cancellable, so in such rings you cannot generally cancel $a$ as you did. See the linked dupe for general results. – Bill Dubuque Nov 30 '21 at 01:25
  • Why reopen this question? I don't know any of the subject matter so I'm curious why this isn't a duplicate. – Toby Mak Dec 19 '21 at 05:07

1 Answers1

0

You indeed get that $a=agc$ for some $g,c\in R$. This does not imply $gc=1$. It does imply that $$a(gc-1)=0.$$ In an integral domain, this implies that either $a=0$ or $gc=1$. In a general commutative ring, it does not.

Note that this also shows that your proof fails for integral domains in case $a=0$. Of course this case is easily taken care of, as then $(b)=(a)=(0)$ and hence $b=0$, so $a=bu$ for any unit $u$.

Servaes
  • 63,261
  • 7
  • 75
  • 163