Ideal is Unit Ideal iff Includes Unity

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

$\blacksquare$