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 $\left({a}\right) \subseteq J$.