Sum of Ideals is Ideal

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.


Then:

$J = J_1 + J_2$ is an ideal of $R$

where $J_1 + J_2$ is as defined in subset product with respect to $\struct{R, +}$.


General Result

Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.


Then:

$J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product with respect to $\struct{R, +}$.


Corollary

$J_1 \subseteq J_1 + J_2$
$J_2 \subseteq J_1 + J_2$


Proof

By definition, $\struct {R, +}$ is an abelian group.

So from Subset Product of Abelian Subgroups, we have that:

$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$

is itself a subgroup of $R$.


Now consider $a \circ b$ where $a \in J, b \in R$.

Then:

\(\ds a \circ b\) \(=\) \(\ds \paren {a_1 + a_2} \circ b\) by definition of $a$
\(\ds \) \(=\) \(\ds \paren {a_1 \circ b} + \paren {a_2 \circ b}\) as $\circ$ is distributive over $+$
\(\ds \) \(\in\) \(\ds J_1 + J_2\) as $\paren {a_i \circ b} \in J_i$ because $J_i$ is an ideal of $R$

Similarly, $b \circ a \in J_1 + J_2$

So by definition $J_1 + J_2$ is an ideal of $R$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Ideals_is_Ideal&oldid=584045"