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:

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

$\blacksquare$