Ideal of Ring is Contained in Radical

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A$ be a commutative ring with unity.

Let $\mathfrak a \subseteq A$ be an ideal.


Then $\mathfrak a$ is contained in its radical:

$\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$


Proof

Let $a \in \mathfrak a$.

By definition of power:

$a^1 = a$

Thus:

$a \in \operatorname{Rad} \left({\mathfrak a}\right)$

$\blacksquare$