# Sum of Ideals is Ideal

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## 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.

### 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.

### 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:

\(\displaystyle a \circ b\) | \(=\) | \(\displaystyle \paren {a_1 + a_2} \circ b\) | by definition of $a$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a_1 \circ b} + \paren {a_2 \circ b}\) | as $\circ$ is distributive over $+$ | ||||||||||

\(\displaystyle \) | \(\in\) | \(\displaystyle J_1 + J_2\) | 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

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Rings: $\S 21$. Ideals: Theorem $40$ - 1972: A.G. Howson:
*A Handbook of Terms used in Algebra and Analysis*... (previous) ... (next): $\S 6$: Rings and fields