Set of non-units in a ring

Question

Let $R$ be a ring with identity. Let ${\rm rad}\: R$ be the radical of $R$, ie the intersection $\bigcap L$ over all maximal left ideals $L$ in $R$. Let $S$ be the set of all non-units in $R$

Question: If $S$ is a left ideal in $R$ then why is $S={\rm rad}\: R$?

I can prove that ${\rm rad}\:R \subset S$ since every maximal ideal of $R$ does not contain any units in $R$. But how do I show that $S \subset {\rm rad}\: R$? Isn't $S$ already a maximal left ideal?

Thanking you in advance.

Answer 1

Hint: Show that if an ideal $I\subset R$ contains a unit, then already $I=R$. Conclude that in your situation $S$ is the unique maximal ideal.