Ideals of Field

Theorem
Let $$\left({R, +, \circ}\right)$$ be a commutative ring 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 $$\left\{{0_R}\right\}$$.

Proof

 * Necessary: 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 $$\left\{{0_R}\right\}$$.


 * Sufficent: 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-zero 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, and the result follows.