Ideals of Field

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Then $\left({R, +, \circ}\right)$ is a field iff the only ideals of $\left({R, +, \circ}\right)$ are $\left({R, +, \circ}\right)$ and $\left\{{0_R}\right\}$.

Necessary Condition
Suppose $\left({R, +, \circ}\right)$ is a field

By definition, a field is a division ring.

So, from Ideals of a Division Ring, if $\left({R, +, \circ}\right)$ is a field then the only ideals of $\left({R, +, \circ}\right)$ are $\left({R, +, \circ}\right)$ and $\left\{{0_R}\right\}$.

Sufficient Condition
Suppose that the only ideals of $\left({R, +, \circ}\right)$ are $\left({R, +, \circ}\right)$ and $\left\{{0_R}\right\}$.

Let $a \in R^*$.

Then $\left\langle {a}\right\rangle$ is a non-null ideal and hence is $R$.

Thus $1_R \in \left\langle {a}\right\rangle$.

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

Therefore $a$ is invertible.

Thus by definition $a$ is a division ring such that $\circ$ is commutative.

The result follows by definition of field.