Principal Ideal is Smallest Ideal
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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: $\S 59.1$