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Let $R$ be a ring with unit and let $M(R)$ be a collection of all non invertible elements in $R$.

Let $R\rhd I$ be a ideal such that $I\neq R$ prove that $I\subseteq M(R)$

I thought to choose some element from $I$ and suppose that the element is invertible, how can I proceed ?

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  • $\begingroup$ Hint: Suppose $I$ contains an invertible element. Try to show that $I$ contains $1$. $\endgroup$ – Kenny Wong Apr 13 '17 at 16:56
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If an ideal contains an invertible element $u$, then it contains $1=uv$, where $v$ is an inverse for $u$.

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