Condition for Ideal to be Total Ring

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$