Condition for Ideal to be Total Ring

This article needs to be linked to other articles. In particular: Links to results used You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. 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 {{MissingLinks}} from the code.

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.