# Condition for Ideal to be Total Ring

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## Theorem

Let $\struct {A, +, \circ}$ 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:

- $\exists C \in A / I: \paren {j + I} \circ C = 1 + I$

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

Then:

\(\ds \paren {j + I} \circ \paren {j' + I}\) | \(=\) | \(\ds 1 + I\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds j \circ j' + I\) | \(=\) | \(\ds 1 + I\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \circ \paren {j \circ j'} + I\) | \(=\) | \(\ds a + I\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists i, i' \in I: \, \) | \(\ds a \circ \paren {j \circ j'} + i\) | \(=\) | \(\ds a + i'\) | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds a \circ j \circ j' + i - i'\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \circ j\) | \(\in\) | \(\ds J\) | Definition of Ideal of Ring | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \circ j \circ j'\) | \(\in\) | \(\ds J\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \cdot j \cdot j' + i - i'\) | \(\in\) | \(\ds J\) | Definition of Ideal of Ring and $I \subsetneq J$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a\) | \(\in\) | \(\ds J\) | 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.

$\blacksquare$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Might be able to get there quicker by using Ideals of Field on the quotient ring $A / I$.Or, using Ideal is Unit Ideal iff Includes Unity You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |