Principal Ideal is Ideal

Theorem
Let $$\left({R, +, \circ}\right)$$ be a ring with unity.

Let $$a \in R$$.

Let $$\left({a}\right)$$ be the ideal of $$R$$ generated by $$a$$.

Then $$\left({a}\right)$$ is a principal ideal if $$\exists a \in R$$ such that $$\left \langle {a} \right \rangle$$ is the ideal generated by $$a$$.

Proof
Let $$a \in R$$.

First we establish that $$\left({a}\right)$$ is an ideal of $$R$$, by verifying the conditions of Test for Ideal.


 * $$a \ne \varnothing$$, as $$1_R \circ a = a \in \left({a}\right)$$.
 * Let $$x, y \in \left({a}\right)$$. Then:

$$ $$ $$ $$


 * Let $$s \in \left({a}\right), x \in R$$.

$$ $$ $$ $$

... and similarly $$s \circ x \in \left({a}\right)$$.

Thus by Test for Ideal, $$\left({a}\right)$$ is an ideal of $$R$$.


 * Now let $$J$$ be an ideal of $$R$$ such that $$a \in J$$.

By the definition of an ideal, $$\forall r \in R: r \circ a \in J$$.

So every element of $$\left({a}\right)$$ is in $$J$$, thus $$\left({a}\right) \subseteq J$$.