If A and B are ideals of a ring, show that A + B = $\{a+b|a \in A, b \in B\}$ is an ideal If A and B are ideals of a ring, show that A + B = $\{a+b|a \in A, b \in B\}$ is an ideal
I have the ideal test but no clue as to what to do with it:


*

*$a-b \in A$ whenever $a,b \in A$

*$ra$ and $ar$ are in A whenever $a \in A$ and $r \in R$

 A: You may be getting confused because you are using $A$, $a$, and $b$ to mean many different things.
We want to show that $A+B$ is an ideal. That is, we want to show that whenever $x,y\in A+B$, then $x-y\in A+B$; and whenever $x\in A+B$ and $r\in R$, we have $rx\in A+B$.
So let us take $x,y\in A+B$. Since $x\in A+B$, we can find $a_1\in A$ and $b_1\in B$ such that $x=a_1+b_1$; and since $y\in A+B$, we can find $a_2\in A$ and $b_2\in B$ such that $y=a_2+b_2$. So then
$$x-y = (a_1+b_1)-(a_2+b_2).$$
Now, can we rewrite it so that it is the sum of something in $A$ and something in $B$? Remember that $A$ and $B$ are ideals.
Likewise, if $x\in A$ and $r\in R$, then we can find $a_1\in A$ and $b_1\in B$ such that $x=a_1+b_1$. Then 
$$rx = r(a_1+b_1).$$
Can you show that this can be written as the sum of something in $A$ and something in $B$?
You also forgot to check whether $xr$ is in $A$ (you did not say your ring is commutative, so you need to check both products). Same idea should work.
A: So, let $a+b, c+d \in A+B$ for $a,c \in A$ and $b, d \in B$. Then $$(a+b)-(c+d)=a+b-c-d=(a-c)+(b-d)\;.$$ Since $A$ and $B$ are ideals then $a-c \in A$ and $b-d \in B$, so $(a+b)-(c+d) \in A+B$.
Then consider $r(a+b)$ for some $r \in R$. Then $r(a+b) = ra+rb$. Since $A$ and $B$ are ideals then $ra \in A$ and $rb \in B$; thus $r(a+b) \in A+B$. Make a similar argument for $(a+b)r$ ...
Therefore, $A+B$ is an ideal.
A: Let $x,y\in A+B$.  We want to show that $x-y\in A+B$.
First, $x = a+b$, for some $a\in A$, $b\in B$, and $y = c+d$, for some $c\in A$, $d\in B$.  Then $x - y = (a-c) + (b-d) \in A+B$, since by assumption, $A$ and $B$ are ideals, and hence $a-c\in A$, for $a,c\in A$, and $b-d\in B$, for $b,d\in B$.
Let $r\in R$, and $x\in A+B$.  Then $x = a+b$ for some $a\in A$, $b\in B$.  Then $rx = ra + rb$, and by assumption, this is in $A+B$, since $ra\in A$, and $rb\in B$.  Similarly, $xr = ar + br \in A+B$.  This proves that $A+B$ is an ideal.
