Proposition

$a \subset (a:b)$
$(a:b)b \subset a$
$((a:b):c) = (a:bc) = ((a:c):b)$
$(\cap_i a_i:b) = \cap_i (a_i:b)$
$(a:\sum_i b_i) = \cap_i (a:b_i)$.

Solution

Lemma

Let $a, b, c$ be ideas. If $\forall x \in a, xb \subset c$, then $ab \subset c$.

(Proof) Let $\sum a_ib_i \in ab$ be given. Then each $a_ib_i \in c$. Since $c$ is closed under addition, $\sum a_ib_i \in c$. Therefore, $ab \subset c$.

1

Let $x \in a$. Then $\forall y \in b, xy \in a$ since $a$ is an ideal. Then $xb \subset a$, so $x \in (a:b)$.

2

For all $x \in (a:b)$, $xb \subset a$. By the Lemma above, $(a:b)b \subset a$.

3

Let $x \in ((a:b):c)$. Then $xc \subset (a:b)$. For all $xz \in xc, (xz)b \subset a$. Therefore, $(xc)b \subset a$ by the Lemma above. Then $x(cb) \subset a$, so $x(bc) \subset a$. Hence, $x \in (a:bc)$.

On the other hand, suppose $x \in (a:bc)$. Then $x(bc) \subset a$. $x(bc) \subset a \implies (xb)c \subset a \implies xb \subset (a:c) \implies x \in ((a:c):b)$.

Therefore, $((a:b):c) = (a:bc)$.

We showed that $((a:b):c) = (a:bc)$. This implies $(a:cb) = ((a:c):b)$. Since $(a:bc) = (a:cb)$, we have $((a:b):c) = (a:bc) = (a:cb) = ((a:c):b)$.

4

For any $x \in A$,

\[\begin{align*} x \in (\cap_i a_i:b) &\iff xb \subset \cap_i a_i \\ &\iff \forall i, xb \subset a_i \\ &\iff \forall i, x \subset (a_i:b) \\ &\iff x \subset \cap_i (a_i:b). \end{align*}\]

5

For any $x \in A$,

\[\begin{align*} x \in (a:\sum_i b_i) &\iff x(\sum_i b_i) \subset a \\ &\implies \forall i, xb_i \subset a \\ &\iff \forall i, x \subset (a:b_i) \\ &\iff x \subset \cap_i(a:b_i). \end{align*}\]

Therefore, it suffices to show that $\forall i, xb_i \subset a \implies x(\sum_i b_i) \subset a$. Let $y_{i_1} + \cdots + y_{i_n} \in \sum_i b_i$ be given where $y_{i_j} \in b_{i_j}$. For each $j$, since $xb_{i_j} \subset a$, $xy_{i_j} \in a$. Since $a$ is closed under finite addition, $xy_{i_1} + \cdots + xy_{i_n} \in a$. Therefore, $\forall i, xb_i \subset a \implies x(\sum_i b_i) \subset a$, so $(a:\sum_i b_i) = \cap_i(a:b_i)$.