"Let $R$ be an arbitrary ring and $\mathfrak a$ an ideal of $R$ admitting an irredundant primary representation $\mathfrak a=\bigcap_{i}\mathfrak q_{i}$ and let $\mathfrak p_i=\sqrt{\mathfrak q_i}$. For a prime ideal $\mathfrak p$ of $R$ to be equal to some $\mathfrak p_i$ it is necessary and sufficient that there exists an element $c\in R\setminus\mathfrak a$ such that the ideal $(\mathfrak a:c)$ is $\mathfrak p$-primary."(Commutative Algebra, Zariski&Samuel).
Why are the prime ideals $\mathfrak p_i$ therefore uniquely determined by $\mathfrak a$?
The property says the following:
Let's take this for granted.
Now suppose that we have another irredundant primary decomposition, $\mathfrak a=\bigcap_j\mathfrak q_j'$, and want to show that $\mathfrak p_j'=\sqrt{\mathfrak q_j'}$ are in fact the same prime ideals as before. In order to prove this it's enough the show that there is $c\in R\setminus\mathfrak a$ such that the ideal $(\mathfrak a:c)$ is $\mathfrak p_j'$-primary. Since the decomposition is irredundant there is $c\in(\bigcap_{k\ne j}\mathfrak q_k')\setminus\mathfrak q_j'$. Now $(\mathfrak a:c)=(\mathfrak q_j':c)$, and this is $\mathfrak p_j'$-primary.