$(R,+,\cdot)$ 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 $(R,+,\cdot)$.
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Hint:
Let $e$ be the identity of the subring. Look at $(1-e)e$.
rschwieb
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umm.. I found the ring R has zero divisors! But I don't know 1' that is subring's identity is zero divisor of R.... – J.G.Lee Mar 24 '14 at 15:52
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@J.G.Lee -- if you compute the value $(1-e)e$ you will have constructed a proof that the identity of the subring is a zero divisor. – Nick Mar 24 '14 at 15:56
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@Nick Thanks!! I'm so stupid;; – J.G.Lee Mar 24 '14 at 16:00
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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 $\,x=1'.$
Bill Dubuque
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