Proving "to contain is to divide" for Dedekind domains

Question

I'm currently reading Number Fields by Marcus and I'm trying to complete a proof left as an exercise. We have the statement as

If A and B are ideals in a Dedekind domain R, then A|B iff A $\supset$ B.

The definition used for Dedekind in the book is:

A Dedekind domain is an integral domain R s.t

(1) Every ideal is finitely generated (2) Every non-zero prime ideal is a maximal ideal (3) R is integrally closed in its field of fractions

K = {$\frac{\alpha}{\beta}:\alpha, \beta \in R, \beta \neq 0$}.

One direction is trivial as it says in the book and it starts off the other direction by saying,

"Assume A $\supset$ B and fix an ideal J s.t AJ is principal, AJ = ($\alpha$). We leave it to the reader to verify that the set C=$\frac{1}{\alpha}$JB is an ideal in R (first show that it is contained in R) and that AC = B".

Now I've tried googling and have found proofs of this statement using different methods but I would really like to see the proof completed in this manner.

I have that since J and B are ideals in R then JB must also be an ideal in R but I don't know how to show that $\frac{1}{\alpha}$ is in R (if it is? I know there is no need for rings to contain multiplicative inverses) or how to show that C is contained in R.

Any help would be much appreciated.

Answer 1

I think after a lot of work, I have part of the answer;

$C=\frac{1}{\alpha}BJ$.

Since $B \subseteq A \Rightarrow BJ \subseteq AJ = (\alpha) \Rightarrow\frac{1}{\alpha}BJ \subseteq R$.

We then have two cases, either JB = AJ $\Rightarrow$ C = R which is an ideal or JB $\subset$ ($\alpha$) $\Rightarrow$ C is a proper ideal of R (not sure if this is true).

Then AC = A($\frac{1}{\alpha}$)JB

= ($\frac{1}{\alpha}$)AJB (since ($\frac{1}{\alpha}$) is just a constant)

= ($\frac{1}{\alpha}$)$\alpha$B

= B $\Rightarrow$ AC = B $\Rightarrow$ A|B.

If anyone could clarify why C is an ideal in the second case that would be class

Answer 2

You showed that $C=\frac{1}{\alpha}JB$ is in $R$, so it remains to show that it is an ideal of $R$. The elements in $C=\frac{1}{\alpha}JB$ are of the form $\frac{1}{\alpha}\sum_{i=1}^n j_i b_i$, where $j_i \in J$ and $b_i \in B$ $\forall i \in \{1, \dots ,n\}$. We have to show that $C$ is an additive group of $R$ and that for each $r \in R$ and $x \in C$, it holds that $rx \in C$.

• Let $x,y \in C$. Then: $$ x-y=\frac{1}{\alpha}\sum_{i=1}^n j_i b_i -\frac{1}{\alpha}\sum_{k=1}^m \tilde{j_k} \tilde{b_k} = \frac{1}{\alpha} \left( \sum_{i=1}^n j_i b_i -\sum_{k=1}^m \tilde{j_k} \tilde{b_k} \right) $$ Since $J$ and $B$ are ideals of $R$, then $JB$ is, so in particular it is an additive subgroup of $R$ and thus closed under substraction, which means that there exist $b'_t \in B$ and $j'_t \in J$ such that: $$ \sum_{i=1}^n j_i b_i -\sum_{k=1}^m \tilde{j_k} \tilde{b_k} = \sum_{t=1}^s j'_t b'_t $$ from where we conclude: $$ x-y = \frac{1}{\alpha} \sum_{t=1}^s j'_t b'_t \in \frac{1}{\alpha}JB $$

• Let $r \in R$ and $x \in C$. Then: $$ rx = r \left(\frac{1}{\alpha}\sum_{i=1}^n j_i b_i \right) = \frac{1}{\alpha}\sum_{i=1}^n j_i (rb_i) $$ Since $B$ is an ideal, we have that $rb_i \in B$ for each $i$, so $rb_i =b'_i$ for some $b'_i \in B$, and thus: $$ rx = \frac{1}{\alpha}\sum_{i=1}^n j_i b'_i \in \frac{1}{\alpha}JB $$