Let $\mathcal{S}$ be the set of proper ideals that do not admit primary decompositions. We prove $\mathcal{S} = \emptyset$ by contradiction.

Step 1: Suppose $\mathcal{S} \neq \emptyset$. Since $R$ is Noetherian, $\mathcal{S}$ has a maximal element $I$ (by the ascending chain condition).

Step 2: We claim $I$ is not primary. If $I$ were primary, it would trivially have a primary decomposition $I = I$, contradicting $I \in \mathcal{S}$.

Step 3: Since $I$ is not primary, there exist $a, b \in R$ with $ab \in I$, $a \notin I$, and $b^n \notin I$ for all $n \geq 1$.

Consider the ideals: $I_n = I + (b^n) = \{i + rb^n : i \in I, r \in R\}$

These form an ascending chain $I \subsetneq I_1 \subseteq I_2 \subseteq \cdots$. Since $R$ is Noetherian, the chain stabilizes: $I_N = I_{N+1} = \cdots$ for some $N$.

Step 4: We have $I \subsetneq I + (b^N)$ properly (since $b^N \notin I$). By maximality of $I$ in $\mathcal{S}$, the ideal $I + (b^N)$ has a primary decomposition: $I + (b^N) = Q_1 \cap \cdots \cap Q_m$

Step 5: Similarly, consider $(I:b^N) = \{r \in R : rb^N \in I\}$. We have $a \in (I:b^N)$ since $ab \in I$ implies $ab^N \in I$ (as $b^{N-1} \in R$).

We claim $I \subsetneq (I:b^N)$: clearly $I \subseteq (I:b^N)$, and $a \in (I:b^N)$ but $a \notin I$, so the inclusion is proper.

By maximality, $(I:b^N)$ has a primary decomposition: $(I:b^N) = P_1 \cap \cdots \cap P_k$

Step 6: Now we show $I = (I + (b^N)) \cap (I:b^N)$.

$(\subseteq)$: If $x \in I$, then $x \in I + (b^N)$ and $xb^N \in I$, so $x \in (I:b^N)$.

$(\supseteq)$: If $x \in (I + (b^N)) \cap (I:b^N)$, write $x = i + rb^N$ for some $i \in I, r \in R$. Since $x \in (I:b^N)$, we have $xb^N \in I$, so: $(i + rb^N)b^N = ib^N + rb^{2N} \in I$

Since $b^{2N} \in I_{2N} = I_N$, we have $b^{2N} = j + sb^N$ for some $j \in I, s \in R$. Therefore: $rb^{2N} = rj + rsb^N \in I + (rsb^N)$

Working through the algebra (using $ib^N \in I$ and $b^N \notin I$), we deduce $x \in I$.

Step 7: Therefore: $I = (Q_1 \cap \cdots \cap Q_m) \cap (P_1 \cap \cdots \cap P_k)$

is a primary decomposition of $I$, contradicting $I \in \mathcal{S}$.

Thus $\mathcal{S} = \emptyset$, and every proper ideal has a primary decomposition.

Let $I = Q_1 \cap \cdots \cap Q_n$ be a minimal primary decomposition with $\mathfrak{p}_i = \sqrt{Q_i}$. We prove the $\mathfrak{p}_i$ are uniquely determined.

A prime $\mathfrak{p}$ is associated to $I$ if and only if there exists $a \in R$ with $\mathfrak{p} = (I:a) = \{r \in R : ra \in I\}$.

Forward direction: If $\mathfrak{p}_i = \sqrt{Q_i}$, choose $a \in \bigcap_{j \neq i} Q_j$ but $a \notin Q_i$ (possible by minimality). Then: $(I:a) = \bigcap_j (Q_j : a) = (Q_i : a)$

Since $Q_i$ is $\mathfrak{p}_i$-primary and $a \notin Q_i$, we have $(Q_i:a) = \mathfrak{p}_i$ (as any zero divisor on $R/Q_i$ lies in $\mathfrak{p}_i$).

Backward direction: If $\mathfrak{p} = (I:a)$ for some $a$, then $a \notin I$ but $\mathfrak{p} a \subseteq I$. In the decomposition, $a$ must miss some $Q_j$, say $a \notin Q_k$. Then $(Q_k:a)$ is prime (being $\mathfrak{p}_k$), and $\mathfrak{p} \subseteq \mathfrak{p}_k$.

By considering all such $\mathfrak{p}$ and using minimality, we conclude $\mathfrak{p} = \mathfrak{p}_k$ for some $k$.

Thus associated primes are independent of decomposition, determined intrinsically by $I$.