# Sum of Ideals

• Jan 19th 2010, 10:07 AM
Ryaη
Sum of Ideals
If A and B are both ideals, then A+B is an ideal. How can you show this?
• Jan 19th 2010, 10:09 AM
Swlabr
Quote:

Originally Posted by Ryaη
If A and B are both ideals, then A+B is an ideal. How can you show this?

What do elements from this ideal look like? Take an arbitrary element of this ideal and multiply it by an arbitrary ring element. Then apply the fact that multiplication in a ring is distributive.
• Jan 19th 2010, 10:27 AM
Ryaη
Is it anything like this?

For some ring R that has A and B as ideals.

$\displaystyle nA \in R$ and $\displaystyle nB \in R$, then $\displaystyle nA+nB \in R$. Thus, $\displaystyle n(A+B) \in R$. Therefore $\displaystyle (A+B) \in R$.
• Jan 19th 2010, 10:43 AM
Swlabr
Quote:

Originally Posted by Ryaη
Is it anything like this?

For some ring R that has A and B as ideals.

$\displaystyle nA \in R$ and $\displaystyle nB \in R$, then $\displaystyle nA+nB \in R$. Thus, $\displaystyle n(A+B) \in R$. Therefore $\displaystyle (A+B) \in R$.

Not quite. Take $\displaystyle x \in A + B$. What does $\displaystyle x$ look like? Then take an arbitrary element from $\displaystyle R$, call it $\displaystyle r$. Then evaluate $\displaystyle xr$ (and $\displaystyle rx$ for completeness).
• Jan 20th 2010, 06:08 PM
Ryaη
I think I got it. Thank you!