Let $D$ be Dedekind domain and $I$ nonempty ideal in $D$. I have to show that there are only finitely many ideals $J$ in $D$ such that $I$ is contained in $J$. My first idea would be:
assume that there are infinitely many such ideals, named $A1, A2, ...$ Let's define $B1:= A1$, $Bn+1 = Bn + An+1$. Can I get contradiction this way (contradiction to the fact that $D$ is Noetherian, thus every ascending sequence od ideals is stationary)?
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Recall that in a Dedekind domain every nonzero ideal is the product of prime ideals in a unique way. Further, recall how "being a subset of" releates to "being a divisor of."
The above does not really answer the question you explcictly asked (though it solves the implicit problem), let me also answer the one you asked more directly:
No, you cannot get a contradiction in this way as the result is not true for arbitrary noetherian domains, you need to use something psecific to Dedekind domains somewhere. Take for example $\mathbb{R}[X,Y]$. In this noetherian domain the ideal generated by $X$ is contained in the one generated by $X$ and $Y-r$ for each $r$.
