For ideals $\mathfrak{a,b}$ if a solution $\mathfrak{c}$ to $\mathfrak{cb}=\mathfrak{a}$ exists, is it unique and equal $(\mathfrak{a}:\mathfrak{b})$?

Question

The motivation for this is to have a better intuition of how to think of the ideal quotient $(\mathfrak{a}:\mathfrak{b})$ as a quotient. Obviously for $a,b \in \mathbb{Q}$, the quotient $\frac{a}{b}$ if it exists is the unique solution $c$ to the equation $cb = a$. So I am curious about how far the analogy can be stretched to ideals.

Question: For ideals $\mathfrak{a,b}$ if a solution $\mathfrak{c}$ to $\mathfrak{cb}=\mathfrak{a}$ exists, is it unique and equal $(\mathfrak{a}:\mathfrak{b})$?

I know that, given ideals $\mathfrak{a,b}$, it does not have to be the case that there is any solution to the equation $\mathfrak{cb} = \mathfrak{a}$, and that all we can say in general is that $(\mathfrak{a}:\mathfrak{b})\mathfrak{b} \subseteq \mathfrak{a}$, with strict inclusion $(\mathfrak{a}:\mathfrak{b})\mathfrak{b} \subsetneq \mathfrak{a}$ being a real possibility. For simplicity I am focusing on the case where we can assume that equality holds, although if you have more to say about the general case I'd be glad to hear it.

I also know that the ideal quotient $\newcommand{\idq}{(\mathfrak{a}:\mathfrak{b})}$$\idq$ is the "largest" ideal $\mathfrak{c}$ such that $\mathfrak{cb} \subseteq \mathfrak{a}$, in the sense that if $\mathfrak{c}$ is an ideal such that $\mathfrak{cb} \subseteq \mathfrak{a}$, then it follows that $\mathfrak{c} \subseteq \idq$. Proof: Let $c \in \mathfrak{c}$ where $\mathfrak{c}$ is such an ideal. Then for all $b \in \mathfrak{b}$, $cb \in \mathfrak{cb} \subseteq \mathfrak{a}$, so $cb \in \mathfrak{a}$ for all $b \in \mathfrak{b}$, or in other words, $c \mathfrak{b} \subseteq \mathfrak{a}$. Then by definition of ideal quotient it follows that $c \in \idq$. Since $c \in \mathfrak{c}$ was arbitrary, $c \subseteq \idq$ follows.

Therefore it's possible to conclude that, if a solution $\mathfrak{c}$ to $\mathfrak{c b} = \mathfrak{a}$ exists, since also $\mathfrak{cb} \subseteq \mathfrak{a}$, we know that $\mathfrak{c} \subseteq \idq$ by the above characterization of $\idq$ . Also, we know that $\idq \mathfrak{b} \subseteq \mathfrak{a} = \mathfrak{cb}$.

Sub-question 1: Is $\idq \mathfrak{b} \subseteq \mathfrak{cb}$ enough to conclude that $\idq \subseteq \mathfrak{c}$ and thus $\mathfrak{c} = \idq$?

If so, then that would guarantee uniqueness of a solution if a solution exists. However, whenever I try to prove that $\idq \mathfrak{b} \subseteq \mathfrak{cb} \implies \idq \subseteq \mathfrak{c}$, I get stuck on the fact that we obviously can't assume that the units of $\mathfrak{b}$ are all units, and thus can't multiply by reciprocals to get the result.

However, if that result actually isn't true, then it might also be the case that the solution of $\mathfrak{cb} = \mathfrak{a}$ isn't unique, with a $\mathfrak{c} \subsetneq \idq$ also solving the equation.

Sub-question 2: If uniqueness isn't valid, then what are some counterexamples of this?

This seems like a really basic question, which is why it bothers me that I'm still stuck on trying to figuring out what the answer actually is, even though I've made some progress.

Answer 1

Here are two kinds of examples.

1. Let the ring be $A = \mathbb{Z}/(2^{4})$ and let $\mathfrak{a} = (0)$ and $\mathfrak{b} = (2^{3})A$. Then for any $i = 1,2,3,4$ the ideal $\mathfrak{c}_{i} = (2^{i})A$ satisfies $\mathfrak{c}_{i}\mathfrak{b} = \mathfrak{a}$ but only $\mathfrak{c}_{1} = (\mathfrak{a}:\mathfrak{b})$. (One can construct similar examples for any nonreduced ring $A$.)
2. Let $k$ be a field and let $A = k[x,y]$ be the polynomial ring. Set $\mathfrak{a} = (x^{3},x^{2}y,xy^{2},y^{3})$ and $\mathfrak{b} = (x,y)$. Then both $\mathfrak{c}_{1} = (x^{2},y^{2})$ and $\mathfrak{c}_{2} = (x^{2},xy,y^{2})$ satisfy $\mathfrak{c}_{i}\mathfrak{b} = \mathfrak{a}$.