A commutative ring with identity is a field if and only it has no nonzero proper ideals

Question

Obviously, if $F$ is a field, and $I$ is it's nonzero ideal, then it contains an invertible element of $F$(any nonzero element of $F$). Denote this element as $a$. Since $I$ is ideal, $aa^{-1} = 1 \in I$. Hence, $I = F$.

But I'm not sure how to prove that any commutative ring with identity without nonzero proper ideals is a field.

Answer 1

Just think backwards:

If you have a commutative ring $R$ with identity, the only missing property to be a field is, that any element is invertible.

So let's assume $R$ is no field. You have some non-invertible element $r \neq 0$ and thus $rR$ is a proper ideal, since $1 \notin rR$.

Answer 2

If $\{0\}$ is the only proper ideal, is a maximal ideal and $R/\{0\}\approx R$ is a field.