# Condition for Ideal to be Total Ring

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## 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.

$\blacksquare$