Assume we have a commutative ring $A$, $a$ is an ideal of $A$. And we have an ideal $b$ proper belongs to $A$. From the definition of ideal, we know that $Aa=a$. I'm now wondering, is the multiplication of ideals $a$ and $b$, which is $ba$, must proper belongs to the ideal $a$?
[See the update for more information.]
Remarks For the things I've already known, one is that the element $1_A$ must not contain in the ideal $b$, neither $a$. So I have a way to the proof is that we can prove $ba$ doesn't contain $1_A$. But since then I didn't got any idea to continue.
Update Since the question seems not clear at the moment, let me briefly explain the context of this question.
As you might already known, Serge Lang introduced the concept of "Dedekind ring" in Chapter II of his famous Algebra, in a quite unusual way (without introduced the concept of Noetherian ring):
And then there is a sequence of exercise in Chapter II about the "Dedekind ring" Lang introduced, one is to prove the unique factorization of ideals. When I look up solutions in Ken Ribet's lecture notes, he mentioned given an maximal ideal $m$ in a Dedekind ring $D$, and an ideal $a$ properly contains in $m$. Then he clarified that because of $m$ is properly belongs to $D$, then $ma$ is properly belongs to $a$. As I lack of knowledge about the Noetherian rings and the well-known definition of Dedekind domains, I couldn't follow up these words.
UPdate I think now I have finally figured out the gap. We just need to multiply both side by $a^{-1}$ and because of $m$ properly belongs to $D$ we can prove the statement. I think I should vote to close this question.