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 principal ideal of $R$ generated by $a$.

Then $\left({a}\right)$ is an ideal of $R$.

Also, if $J$ is an ideal of $R$ and $a \in J$, then $\left({a}\right) \subseteq J$.

That is, $\left({a}\right)$ is the smallest ideal of $R$ to which $a$ belongs.

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


 * $\left({a}\right) \ne \varnothing$, 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 $R$.

Proof that Principal Ideal is Smallest
Let $J$ be an ideal of $R$ such that $a \in J$.

By the definition of an ideal, $\forall r,s \in R: r \circ a \circ s \in J$. Also, $J$ is a group under $+$.

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