Let be $D$ a commutative domain, $\mathfrak{a,b,c}\subseteq D$ ideals. Show that:
if $\mathfrak{a\subseteq b}$ and $\mathfrak{bc}=\lambda D$ then there is an ideal $\mathfrak{b}'$ such $\mathfrak{a=bb'}$
How to work this?
My work:
Set $\mathfrak{b}'=\{d\in D: bd\in\mathfrak{a}, \forall b\in\mathfrak{b}\}$. is clear to see that $\mathfrak{b}'$ is and ideal and $\mathfrak{b \subseteq a} $ but i thinks that I'm missing something because not using the fact that $\mathfrak{b}$ is invertible ($\mathfrak{bc}=\lambda D$)
What is the idea behind this exercise? I'm in a number theory course and we have make properties that intuition tells me that ideals works as numbers but I can't figure out the intuition behind this problem.
okey. I've worked and found the solution:
$$\mathfrak{ac\subseteq bc=}\lambda D$$ * then exists some $\mathfrak{b}'$ ideal such $\mathfrak{ac}=\lambda \mathfrak{b}'$ since $D$ is commutative, $$\mathfrak{abc}=\lambda\mathfrak{ b b}'$$ $$\mathfrak{a}\lambda D=\lambda\mathfrak{bb'}\Rightarrow \mathfrak{a}D=\mathfrak{a=bb'}$$
(*) Let be $D$ a commutative domain, $\lambda\in D\setminus \{0\}$ and $A\subset D$ an ideal. Then
Proof $$A' = \{ x \in D : \lambda x \in A\}.$$
Now it remains to see that $A'$ has the required properties. We need to verify $A = \lambda A'$. By definition, we have $\lambda A' \subset A$. And since by assumption $A \subset \lambda D$, for every $a\in A$ there is a $d\in D$ with $a = \lambda d$. That implies $\lambda d \in A$, and hence $d \in A'$, so we also have $A \subset \lambda A'$.