TITLE: Let R be a commutative ring and let $A$ and $B$ be ideals of $R$. Show that if $A + B = R$, then $AB = A\cap B$
QUESTION [5 upvotes]: Let $R$ be a commutative ring and let $A$ and $B$ be ideals of $R$. 
(i) Show that if $A + B = R$, then $AB = A\cap B$
(ii) Now let $R$ be a Euclidean domain. prove that if $AB = A\cap B$, then $A + B = R$
I can define 
$$
A + B = \{ a + b : a\in A; b \in B\}
$$
$$
AB = \{a_1b_1 + \ldots +a_nb_n : n\in \mathbb{N}; a_i\in A; b_j\in B\}
$$
I can only prove that $AB$ is also an ideal in $R$.

REPLY [2 votes]: For every $t$ $\in$ $A \cap B$, $t=a+b$, $a \in A$ ,$b \in B$.Then we can get $a \in B$ ,$b \in A$ from $t$ $\in$ $ A \cap B$ And $1=c+d$, $c \in A$, $d \in B$. So $t=1t=(c+d)(a+b)=ca+cb+da+db \in AB$.