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Problem statement:

Let $c \in R$ and let $ I = \{rc\mathrel{|}r\in R\} $. If $R$ is commutative but has no identity, is $c$ an element of the ideal $I$?

Proof:

Suppose $c \notin I$ and $R$ is a commutative ring with identity. Then $\exists a\in I$ s.t. $a=rc$, where $r\in R$. Then, since $R$ has identity $\exists r^{-1} \in R$ s.t. $rr^{-1}c\in I$ since $I$ is ideal. Therefore, $rr^{-1}c = 1_{R} \cdot c = c\in I $. Hence by contraposition if $R$ is commutative without identity, then $ c\notin I$.

Does this work?

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The plan of your proof is to show that if $R$ has no identity element then $I = Rc$ cannot contain $c$. This is clearly wrong if $c = 0$. For a counter-example with $c \neq 0$, consider the ring $R = 2\Bbb{Z} \times \Bbb{Z}$ comprising pairs $(i, j)$ of integers where $i$ is even, with addition and multiplication carried out element-wise: $(i, j) + (i', j') = (i + i', j + j'),\,$ $(i, j)(i', j') = (ii', jj').\,$ $R$ has no multiplicative identity: there is no element $(i, j) \in R$ such that $(i, j)(2, 1) = (2, 1).\,$ However, if $c = (0, 1),\,$ $c$ belongs to the ideal $I$ generated by $c$, because $c = (0, 1)c$.

Your proof goes wrong because you are assuming that if $R$ has no identity, then the equation $xc = c$ has no solutions for any $c\neq 0$, but it may have solutions for some $c$ but not for all $c$.

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