1

Let $R$ be a commutative ring with unit.

On the elements of $R$ we can define the following relations:$$a\sim_1b\iff a=ub\text{ for some unit }u$$

and:$$a\sim_2 b\iff r,s\in R\text{ exist with }a=rb\text{ and }b=sa$$

It is clear to me that both relations are equivalence relations and that $a\sim_1b\implies a\sim_2b$.

Also it is clear to me that the opposite of this is true if $a$ or $b$ is not a zero-divisor.

Questions:

1) Do we always have $a\sim_2b\implies a\sim_1b$? (I suspect not)

2) If not then can you provide a counterexample?

3) Is the condition "$R$ has no zero-divisors" (sufficient for $a\sim_2b\implies a\sim_1b$) also a necessary condition for $a\sim_2b\implies a\sim_1b$?

Thank you in advance and sorry if this question is a duplicate.

Vera
  • 3,469

1 Answers1

1

As you suspect, the answer to your first question is no. A counterexample is $R=\mathbb{Z}[X,Y,Z]/(X(1-YZ))$.

One may check that the units of $R$ are $\pm 1_R$. I let you find two elements generating the same ideal but which do not differ by a unit.

For your last question, the answer is NO as well. The ring $\mathbb{Z}/4\mathbb{Z}$ has zero divisors, but two elements generating the same ideal necessarily differ by a unit.

GreginGre
  • 15,028
  • Thank you. In the meanwhile I found this question, showing that my question is actually a duplicate. One question though: is $\mathbb Z/4\mathbb Z$ an exception or is it true that for every ring $\mathbb Z/n\mathbb Z$ the relations coincide? I realize that this is a question that essentially differs from the one I posted, so do not feel obliged to answer it. – Vera Aug 20 '19 at 11:22
  • I kinda remember that the same phenomenon holds for $\mathbb{Z}/{n\mathbb{Z}}$. I have seen the proof somewhere but cannot find it where for the moment. If I do, I'll post a proof or a link. – GreginGre Aug 20 '19 at 11:41
  • Thank you very much. – Vera Aug 20 '19 at 12:13