Condition for Ideal to be Total Ring

Theorem
Let $\left({A, +, \circ}\right)$ be a commutative ring with unity.

Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.

Let $J$ be an ideal of $A$ such that $I \subsetneq J$.

Then:
 * $A = J$

Proof
Let $A$ be a commutative ring with unity.

Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.

Let $J$ be an ideal of $A$ such that $I \subsetneq J$.

From Ideal is Subring:
 * $J \subseteq A$

It remains to be proved that that $A \subseteq J$.

Let $a \in A$.

As $I \subsetneq J$, it follows from definition of proper subset that:


 * $\exists j \in J: j \notin I$

Consider the coset $j + I \in A / I$.

As $A / I$ is a field:


 * $\exists C \in A / I: \left({j + I}\right) \circ C = 1 + I$

Let $j' \in A$ be such that $C = j' + I$.

Then:

By definition of subset:
 * $A \subseteq J$

Thus $A = J$ follows by Equality of Sets.