If $p=pc$, does this imply $c=1$ in ring $R$?

Question

Let $R$ be a ring, where $p=pc$ for $p,c\in R$. Does $c$ have to be equal to $1$? $R$ may not be an integral domain.

Note that $p,c\neq 0$.

Thanks!

Answer 1

In $\;R=\Bbb Z_8:=\Bbb Z/8\Bbb Z\;$ ,we have that

$$6=6\cdot 5$$

but certainly $\;6\neq 1\pmod 8\;$ ...

Answer 2

Hint $\ \Bbb Z[p,c]/(p-pc)$ $\phantom{......................}$

Answer 3

Nontrivial idempotents give counterexamples. The simplest one that comes to mind is $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ in $M_2(\mathbb{Z})$, for instance.

Answer 4

An integral domain is defined to be a ring with the cancellation property. So cancellation does not work in general in a ring that is not an integral domain. Zero divisors, which exist in any ring that is not an integral domain, will provide counterexamples. Consider $R^2$ as a direct product of any unital ring $R$ with itself. Addition and multiplication are defined component-wise as in $R$. Take $p=(a,0)$, for any nonzero a $ \epsilon $ $R$. Then take $c=(1,b)$ for any b $\epsilon$ $R$. Then $pc=(a,0)(1,b)=(a,0)=p$. So no, it does not hold. It does most certainly hold in integral domains, however.