I would really like to see a simple proof for the following question, if possible.
Let $R$ be a Dedekind domain. Then, $xy \in (x^2,y^2)R$ for any $x,y$ in $R$. Also, show that this fails in general.
Also, I'm pretty sure this generalizes to the fact that $(x_{1}\ldots x_{n+1})^{n} \in (x_{1}^{n+1}, \ldots,x_{n+1}^{n+1})R$ if $R$ is a Dedekind domain. Again, a proof would be really nice to see.
Thanks in advance!
Hint $\ $ It's a one-line proof: cancel $\rm\:(x,y)\:$ from $\rm\:(x,y)\,(x^2,y^2)\, =\, (x,y)^3$
The generalization to higher powers and multiple generators follow similarly, for example see my post on the Freshman's Dream $\rm\ (A+B)^n = A^n\! + B^n\:$ for GCDs and invertible ideals.
Remark $\ $ Conversely, an integrally closed domain satisfying $\rm\:xy \in (x^2,y^2)\:$ is Prüfer, i.e. finitely generated nonzero ideals are invertible. This holds true more generally if $\rm\ x^{n-1}y \in (x^n,y^n)\:$ for some $\rm\:n>1,\:$ or if $\rm\:(x,y)^n = (x^n,y^n)\:$ for some $\rm\:n>1.\:$
Prüfer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfying the Chinese Remainder Theorem for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C,\ $ or $\rm\ (A + B)\ (A \cap B) = A\ B,\ $ or $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm\:A\:$ etc. It has been remarked that there are probably around $100$ such characterizations known. See this answer for around $30$ characterizations.
By unique factorization of ideals, let $(x) = \prod_{i} P^{\alpha_i}_i$ and $(y) = \prod_i P^{\beta_i}_i$. Then $(x^2, y^2) = \prod_i P^{\min\{2\alpha_i, 2\beta_i\}}_i = \prod_i P^{2\min\{\alpha_i, \beta_i\}}_i$, and $(xy) = \prod_i P^{\alpha_i + \beta_i}_i$. Since $\alpha_i + \beta_i \geq 2\min\{\alpha_i, \beta_i\}$, you have $(xy) \subset (x^2,y^2)$.
