Ideal is Unit Ideal iff Includes Unity
Jump to navigation
Jump to search
Theorem
Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
- $\mathfrak a = A \iff 1 \in \mathfrak a$
where $1$ denotes the unity of $A$.
Proof
$\implies$ is trivial.
To see $\impliedby$, suppose $1 \in \mathfrak a$.
Let $a \in A$ be arbitrary.
Then:
\(\ds a\) | \(=\) | \(\ds a 1\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(\in\) | \(\ds \mathfrak a\) | since $1 \in \mathfrak a$, by Definition of Ideal of Ring |
$\blacksquare$