# ideal without invertible elements

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 ?

• Hint: Suppose $I$ contains an invertible element. Try to show that $I$ contains $1$. – Kenny Wong Apr 13 '17 at 16:56

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