Condition for Ideal to be Total Ring
In particular: Links to results used.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can 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.
$\blacksquare$
Might be able to get there quicker by using Ideals of Field on the quotient ring $A / I$.
You can help Proof Wiki by redesigning it.
To discuss this proof in more detail, feel free to use the talk page.
If you are able to fix this proof, then when you have done so you can remove this instance of {{Improve}} from the code.
If you would welcome a second opinion as to whether your improvement is correct, add an invitation to {{Proofread}} the page (see the proofread template for usage).