ideal without invertible elements

Question

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 ?

Answer 1

If an ideal contains an invertible element $u$, then it contains $1=uv$, where $v$ is an inverse for $u$.