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