Cancellation law of ideals in a certain ring

Question

Let $R$ be a integral domain satisfying the following property.

For any non-zero ideal $A$ of $R$, there exist $a \in R\ (a \neq 0)$ and a non-zero ideal $B$ of $R$ such that $AB=(a)$.

Let $A,B,C$ be non-zero ideals of $R$. I can prove that

$AB=AC \Rightarrow B=C$.

I then want to prove the following:

$B \subset A \Rightarrow$ there exists a non-zero ideal $C$ of $R$ such that $B=AC$.

I got a hint as follows: There exists an ideal $J$ and $a \ne 0$ such that $AJ=(a)$. Then since $JB\subset AJ$ we have that $C=(1/a)JB$ is an ideal in $R$. Lastly, note we have $AC=B.$ Shubhodip Mondal's answer helps me to understand that the hint is nothing but a proof.

My additional question is to prove

Every non-zero prime ideal is a maximal ideal.

Actually if $R$ is a Dedekind domain then every non-zero prime ideal is principal hence maximal. For instance see proof of Theorem 3.32 here. But just given a property of $R$ in the beginning, how can we prove that?

Any help would be much appreciated.

Answer 1

$AJ = ( \alpha)$.

$B \subseteq A$. So for $b_i \in B$ and $j_i \in J$ , $ \sum b_i j_i \in AJ = (\alpha) $.

So, $\frac{1}{\alpha} \sum b_i j_i \in R$.

Therefore, $\frac 1 \alpha BJ$ happens to be an ideal of $R$ and you do not need anywhere that $\alpha$ is an unit.

Answer 2

Actually it's very easy to show that every non-zero prime ideal is maximal.

Let $P$ be a non-zero prime ideal which is not maximal and $M\supsetneq P$ a maximal ideal. Then, there is a non-zero ideal $C$ such that $P=CM$. From the definition of prime ideals we get $P\supseteq C$, so $C=P$. Thus $PR=P=PM$ which gives $M=R$, a contradiction.