# Commutative and Unitary Ring with 2 Ideals is Field

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$.

Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are:

- $\set {0_R}$

and: $\struct {R, +, \circ}$ itself.

That is, such that $\struct {R, +, \circ}$ has no non-null proper ideals.

Then $\struct {R, +, \circ}$ is a field.

## Proof

From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are ideals of $\struct {R, +, \circ}$.

Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.

Let $\ideal a$ be the principal ideal of $R$ generated by $a$.

We have that $\ideal a$ is a non-null ideal and hence $\ideal a = R$.

Thus $1_R \in \ideal a$.

Thus $\exists x \in R: x \circ a = 1_R$ by the definition of principal ideal.

Therefore $a$ is invertible.

As $a$ is arbitrary, it follows that all such $a$ are invertible.

Thus by definition $\struct {R, +, \circ}$ is a division ring such that $\circ$ is commutative.

The result follows by definition of field.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Rings: $\S 21$. Ideals: Theorem $37$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $9$: Rings: Exercise $6 \ \text {(i)}$