# 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:

1. $a-b \in A$ whenever $a,b \in A$
2. $ra$ and $ar$ are in A whenever $a \in A$ and $r \in R$
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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.

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+ 0.7 for the first sentence. + 0.3 for the rest :) – The Chaz 2.0 May 8 '12 at 3:46
I think you meant $x-y=(a_1+b_1)-(a_2+b_2)$ – Joseph Skelton May 8 '12 at 3:51
@JosephSkelton: Thank you; indeed I did. – Arturo Magidin May 8 '12 at 3:52

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.

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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.

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