I understand that if $I$ is a maximal ideal of $R$, then $I+xR =R$ $\forall x \notin I$, but I fail to grasp how the converse can be true. Would anyone be so kind as to offer me a proof/ some hints at how the proof can be done?
2026-03-28 15:26:59.1774711619
Maximal Ideal and its relationship with the ring
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The converse is obviously true. Since if $I$ satisfies the condition then $I$ cannot be contained in any proper ideal $J$ of $R$ because if $x\in J\setminus I$ then $R=Rx +I\subseteq J+I\subseteq J+J =J$.