Question: Suppose that M is a proper ideal in a ring R. We proved in class that if R/M is simple, then Al is a maximal ideal in R. Give a careful proof of the converse direction. That is, prove that if M is maximal. then R/M is simple.
Remark: Remember that M is the zero element of R/M. Therefore the smallest ideal in R/M is {M}, the set that contains only the zero element.
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