Let $R$ be a commutative ring with 1, and let $A,B,C \subseteq R$ be ideals. Recall that $A+B = \{a+b \ |\ a \in A, b \in B\}$. Prove the following.

1. $A+B$ is an ideal
2. $A,B \subseteq A+B$
3. If $A,B \subseteq C$ then $A+B \subseteq C$
4. $A+B = (A,B)$
5. If $(A,B) = (1)$, then there exist $\alpha \in A$ and $\beta \in B$ such that $\alpha+\beta = 1$.
6. If $(A,B) = (1)$ and $BC \subseteq A$, then $C \subseteq A$.

Suppose $a_1+b_1,a_2+b_2 \in A+B$, and let $r \in R$. Then $(a_1+b_1) + r(a_2+b_2) = (a_1+ra_2) + (b_1+rb_2) \in A+B$ since $A$ and $B$ are ideals. Moreover, $0 = 0+0 \in A+B$. By the submodule criterion, $A+B \subseteq R$ is an ideal.

For all $a \in A$, $a = a+0 \in A+B$. So $A \subseteq A+B$, and similarly $B \subseteq A+B$.

Suppose $A, B \subseteq C$ for some ideal $C$. Since $C$ is an ideal, it is closed under sums, so that $a+b \in C$ for all $a \in A$ and $b \in B$. Thus $A+B \subseteq C$.

Note that $A, B \subseteq (A,B)$, so that $A+B \subseteq (A,B)$ by the previous point. Now let $x \in (A,B)$; by definition, $x = \sum r_ic_i$ for some $r_i \in R$ and $c_i \in A \cup B$. Collecting terms in $A$ and $B$, we have $x = \alpha + \beta$ for some $\alpha \in A$ and $\beta \in B$. Thus $A+B = (A,B)$.

Suppose $(A,B) = (1)$. By the previous point, $A+B = (1)$, and in particular $1 = \alpha+\beta$ for some $\alpha \in A$ and $\beta \in B$.

Suppose $(A,B) = (1)$. By the previous point, there exist $\alpha \in A$ and $\beta \in B$ such that $\alpha+\beta = 1$. Now let $\gamma \in C$. We have $\gamma = \gamma(\alpha+\beta) = \gamma\alpha + \gamma\beta$. Since $BC \subseteq A$, $\gamma\beta \in A$, and so $\gamma \in A$. Thus $C \subseteq A$.