Prove that a proper ideal of R is a maximal ideal of R, with R being a commutative ring with unity

Question

R is a commutative ring with unity and A is a proper ideal of R, and every element of R that is not in A is a unit of R. Prove that A is a maximal ideal of R.

Can anyone give a head start on how to approach this problem? I only know I'm supposed to find a B=A or B=R with $A\subseteq B\subseteq R$.

Answer 1

Hint: If $B \supsetneq A$, $B$ contains a unit. By absorption, show that $1 \in B$. By absorption again, show $R \subseteq B$.

Answer 2

If a unit belongs to an ideal $I$ then $1\in I$ and so the ideal is the whole ring. So proper ideals have to be contained in the subset of non-units. Your hypothesis says the ideal consists of all the non-units, so it has to be a maximal ideal. (Actualy in such a case it would be a unique maximal ideal; such rings are called local rings.)

Answer 3

Since everything in the set $R\setminus A$ has an inverse, all nonzero elements of the quotient ring $R/A$ have inverses.

So $R/A$ is a field, and that means $A$ is a maximal ideal.