# 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:

 $\ds \left({j + I}\right) \circ \left({j' + I}\right)$ $=$ $\ds 1 + I$ $\ds \implies \ \$ $\ds j \circ j' + I$ $=$ $\ds 1 + I$ $\ds \implies \ \$ $\ds a \circ \left({j \circ j'}\right) + I$ $=$ $\ds a + I$ $\ds \implies \ \$ $\, \ds \exists i, i' \in I: \,$ $\ds a \circ \left({j \circ j'}\right) + i$ $=$ $\ds a + i'$ $\ds \implies \ \$ $\ds a$ $=$ $\ds a \circ j \circ j' + i - i'$ $\ds \implies \ \$ $\ds a \circ j$ $\in$ $\ds J$ Definition of Ideal of Ring $\ds \implies \ \$ $\ds a \circ j \circ j'$ $\in$ $\ds J$ $\ds \implies \ \$ $\ds a \cdot j \cdot j' + i - i'$ $\in$ $\ds J$ Definition of Ideal of Ring and $I \subsetneq J$ $\ds \implies \ \$ $\ds a$ $\in$ $\ds J$ Definition of Ideal of Ring and $I \subsetneq J$

By definition of subset:

$A \subseteq J$

Thus $A = J$ follows by definition of set equality.

$\blacksquare$