Ring and Subring have different identity

Question

$(R,+,\cdot)$ is a commutative ring with identity 1 and $(S,+,\cdot)$ is a subring with identity 1'. Prove that if $1\ne 1'$, then 1' is a zero divisor in $(R,+,\cdot)$.

Answer 1

Hint:

Let $e$ be the identity of the subring. Look at $(1-e)e$.

Answer 2

Hint $\ $ The quadratic $\,x^2 = x\,$ has three solutions $\,0,1,1'$ hence the difference of some pair of roots must be a zero divisor. Indeed, evaluate $\,x(x-1)=0\,$ at $\,x=1'.$