If $A\subset \lambda D$ then $\exists A'$ such $A=\lambda A'$

Question

Let be $D$ a commutative domain, $\lambda\in D\setminus \{0\}$ and $A\subset D$ an ideal. Then

if $A\subset \lambda D$ then exists $A'\subset D$ such $A=\lambda A'$

my work: If $A=\lambda D$ since $A$ is an ideal then $D$ is and ideal and is enough take $A'=D$. If $A\neq D$ then $\exists b\in D:\lambda b\notin A\rightarrow b\notin A$. If we take $B=\{b\in D: \lambda b \notin A\}$ then we have $A'=D\setminus B.$ am I right?

Any hint? Any recomended book for theory? Please

Answer 1

Your definition of $A'$ is a little roundabout. It's more direct to define

$$A' = \{ x \in D : \lambda x \in A\}.$$

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 $A \subset \lambda A'$.

Although it is not explicitly stated, I think that one is also expected to show that $A'$ is an ideal of $D$. That means one needs to verify

• $0 \in A'$,
• $x,y \in A' \implies (x-y) \in A'$,
• $x\in A',\, \mu \in D \implies \mu x \in A'$.

All these verifications are straightforward (the last, since $D$ is commutative).