# Principal Ideal is Ideal

## Theorem

Let $\struct {R, +, \circ}$ be a ring with unity.

Let $a \in R$.

Let $\ideal a$ be the principal ideal of $R$ generated by $a$.

Then $\ideal a$ is an ideal of $R$.

## Proof

First we establish that $\ideal a$ is an ideal of $R$, by verifying the conditions of Test for Ideal.

$\ideal a \ne \O$, as $1_R \circ a = a \in \ideal a$.

Let $x, y \in \ideal a$.

Then:

 $\ds \exists r, s \in R: \ \$ $\ds x$ $=$ $\ds r \circ a, y = s \circ a$ $\ds \leadsto \ \$ $\ds x + \paren {-y}$ $=$ $\ds r \circ a + \paren {-s \circ a}$ $\ds$ $=$ $\ds \paren {r + \paren {-s} } \circ a$ $\ds \leadsto \ \$ $\ds x + \paren {-y}$ $\in$ $\ds \ideal a$

Let $s \in \ideal a, x \in R$.

 $\ds s$ $\in$ $\ds \ideal a, x \in R$ $\ds \leadsto \ \$ $\ds \exists r \in R: s$ $=$ $\ds r \circ a$ $\ds \leadsto \ \$ $\ds x \circ s$ $=$ $\ds x \circ r \circ a$ $\ds$ $\in$ $\ds \ideal a$

and similarly $s \circ x \in \ideal a$.

Thus by Test for Ideal, $\ideal a$ is an ideal of $R$.

$\blacksquare$