Product of two ideals in Dedekind domains

Question

Let $\mathcal{O}$ be a Dedekind domain and $I=(x_1,\ldots, x_n),J=(y_1,\ldots, y_m)\subseteq \mathcal{O}$ two ideals. Is it possible, that $IJ\neq K$ with $K$ the ideal generated by the products $x_iy_j$?

For PID this is obvious and since a Dedekind Domain is PID iff it is UFD, most of the examples I have in mind don't work to disprove it...

Answer 1

Since $x_iy_j\in IJ$ for all $i,j$, and these generate $K$, we have $K\subseteq IJ$.

Since everything in $IJ$ is of the form $\sum a_ib_i$ with $a_i\in I, b_i\in J$, you can just rewrite the $a_i$ and $b_i$ in terms of the $x_i$ and $y_j$ , distribute everything, and wind up with a linear combination of $x_iy_j$. Thus $IJ\subseteq K$.

Answer 2

It is an immediate inductive consequence of the Ideal Distributive Law $\, A(B+C) = AB + AC,\ $ which is true in any ring.

While you are thinking about the relationship with PIDs, note that the distributive law also holds for gcds. Therefore, by exploiting their common arithmetical laws, one can often give unified proofs that work for both gcds and ideals. For example, see this proof of the *Freshman's Dream for ideals/gcds $\,(A+B)^n = A^n+B^n,\,$ or $\,(a,b)^n = (a^n,b^n).$