Principal Ideal is Smallest 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 $a$.


Let $J$ be an ideal of $R$ such that $a \in J$.

Then $\ideal a \subseteq J$.

That is, $\ideal a$ is the smallest ideal of $R$ to which $a$ belongs.


Proof

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 $\ideal a$ is in $J$.

Thus $\ideal a \subseteq J$.

$\blacksquare$


Sources

  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 59.$ Principal ideals in a commutative ring with a one: $\S 59.1$
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