Let $A$ be a Noetherian integral domain whose Krull dimension equals 1. Given $I\subseteq A$ any ideal, prove or give a counter example: the minimal primary decomposition of $I$ is unique.
If $I = \cap_{i=1}^n q_i = \cap_{i=1}^m Q_i,$ with $q_i,Q_i$ primary and $r(q_i)=p_i,r(Q_i)=P_i$, since $I\subseteq r(I)$, then
$$ \cap q_i \subseteq \cap P_i.$$
So, given $j$, it follows that $\cap q_i\subseteq P_j$ and therefore, since $P_j$ is prime, exists $i$ such that $q_i\subset P_j$. Now $p_i=r(q_i)$ is the smallest prime containing $q_i$, so $p_i=P_j$. Since $\{0\}$ is prime, it follows that $P_j$ is maximal and hence $p_i=P_j$.
I cant conclude anything from this...