[proofplan] We reduce to the local case by localising at $\mathfrak{p}$, which preserves both the Noetherian property and the number of generators. In the local ring $A_\mathfrak{p}$, the minimality of $\mathfrak{p}$ over $\mathfrak{a}$ forces $\mathfrak{a} A_\mathfrak{p}$ to be $\mathfrak{p} A_\mathfrak{p}$-primary (by the [Characterisation of $\mathfrak{m}$-Primary Ideals](/theorems/2900)). Since $\mathfrak{a} A_\mathfrak{p}$ is generated by $r$ elements, the $\delta$-invariant satisfies $\delta(A_\mathfrak{p}) \le r$. The [Dimension Theorem](/theorems/860) then gives $\operatorname{ht}(\mathfrak{p}) = \dim(A_\mathfrak{p}) = \delta(A_\mathfrak{p}) \le r$. [/proofplan] [step:Localise at $\mathfrak{p}$ to reduce to the local case] Let $\mathfrak{p}$ be a minimal prime ideal over $\mathfrak{a}$. Consider the localisation $A_\mathfrak{p} = (A \setminus \mathfrak{p})^{-1} A$. This is a Noetherian local ring with maximal ideal $\mathfrak{p} A_\mathfrak{p}$. The ideal $\mathfrak{a} A_\mathfrak{p}$ is generated by the images $x_1/1, \ldots, x_r/1$ of the generators of $\mathfrak{a}$, so $\mathfrak{a} A_\mathfrak{p}$ is generated by at most $r$ elements. By the standard identity $\operatorname{ht}(\mathfrak{p}) = \dim(A_\mathfrak{p})$ (the height of $\mathfrak{p}$ in $A$ equals the Krull dimension of the local ring $A_\mathfrak{p}$), it suffices to show $\dim(A_\mathfrak{p}) \le r$. [guided] The height $\operatorname{ht}(\mathfrak{p})$ is defined as the supremum of lengths of chains of prime ideals $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_d = \mathfrak{p}$ descending from $\mathfrak{p}$. This coincides with $\dim(A_\mathfrak{p})$ because localisation at $\mathfrak{p}$ establishes a bijection between the set of prime ideals of $A$ contained in $\mathfrak{p}$ and the set of all prime ideals of $A_\mathfrak{p}$, preserving inclusion order. The localisation $A_\mathfrak{p}$ is Noetherian because $A$ is Noetherian and localisation preserves the Noetherian property. It is local with maximal ideal $\mathfrak{p} A_\mathfrak{p}$. The ideal $\mathfrak{a} A_\mathfrak{p}$ is the extension of $\mathfrak{a}$ to $A_\mathfrak{p}$ and is generated by $x_1/1, \ldots, x_r/1$ (the images of the generators under the localisation map $A \to A_\mathfrak{p}$). [/guided] [/step] [step:Show $\mathfrak{a} A_\mathfrak{p}$ is $\mathfrak{p} A_\mathfrak{p}$-primary by exploiting minimality of $\mathfrak{p}$] We show $\sqrt{\mathfrak{a} A_\mathfrak{p}} = \mathfrak{p} A_\mathfrak{p}$. Let $\mathfrak{q}$ be any prime ideal of $A_\mathfrak{p}$ containing $\mathfrak{a} A_\mathfrak{p}$. The contraction $\mathfrak{q}^c = \mathfrak{q} \cap A$ is a prime ideal of $A$ satisfying: \begin{align*} \mathfrak{a} \subset \mathfrak{q}^c \subset \mathfrak{p}, \end{align*} where the first inclusion holds because $\mathfrak{a} A_\mathfrak{p} \subset \mathfrak{q}$ implies $\mathfrak{a} \subset \mathfrak{q}^c$, and the second holds because every prime of $A_\mathfrak{p}$ contracts to a prime contained in $\mathfrak{p}$. Since $\mathfrak{p}$ is a minimal prime over $\mathfrak{a}$ and $\mathfrak{q}^c$ is a prime containing $\mathfrak{a}$ with $\mathfrak{q}^c \subset \mathfrak{p}$, minimality forces $\mathfrak{q}^c = \mathfrak{p}$. Therefore $\mathfrak{q} = \mathfrak{p} A_\mathfrak{p}$ (since the extension of $\mathfrak{q}^c = \mathfrak{p}$ to $A_\mathfrak{p}$ is $\mathfrak{p} A_\mathfrak{p}$, and $\mathfrak{q}$ is the extension of its contraction for primes of $A_\mathfrak{p}$). Since every prime of $A_\mathfrak{p}$ containing $\mathfrak{a} A_\mathfrak{p}$ equals $\mathfrak{p} A_\mathfrak{p}$, we have $\sqrt{\mathfrak{a} A_\mathfrak{p}} = \mathfrak{p} A_\mathfrak{p}$. By the [Characterisation of $\mathfrak{m}$-Primary Ideals](/theorems/2900), the ideal $\mathfrak{a} A_\mathfrak{p}$ is $\mathfrak{p} A_\mathfrak{p}$-primary. [guided] The key step is to show that $\mathfrak{p} A_\mathfrak{p}$ is the only prime of $A_\mathfrak{p}$ that contains $\mathfrak{a} A_\mathfrak{p}$. This is where the minimality of $\mathfrak{p}$ over $\mathfrak{a}$ is used. Let $\mathfrak{q}$ be any prime of $A_\mathfrak{p}$ with $\mathfrak{a} A_\mathfrak{p} \subset \mathfrak{q}$. The primes of $A_\mathfrak{p}$ are in bijection with the primes of $A$ contained in $\mathfrak{p}$: every prime $\mathfrak{q}$ of $A_\mathfrak{p}$ has the form $\mathfrak{q} = \mathfrak{q}^c A_\mathfrak{p}$ where $\mathfrak{q}^c = \mathfrak{q} \cap A$ is a prime of $A$ with $\mathfrak{q}^c \subset \mathfrak{p}$. The condition $\mathfrak{a} A_\mathfrak{p} \subset \mathfrak{q}$ implies $\mathfrak{a} \subset \mathfrak{q}^c$ (contract both sides to $A$). So $\mathfrak{q}^c$ is a prime of $A$ satisfying $\mathfrak{a} \subset \mathfrak{q}^c \subset \mathfrak{p}$. But $\mathfrak{p}$ is minimal over $\mathfrak{a}$, meaning there is no prime strictly between $\mathfrak{a}$ and $\mathfrak{p}$ containing $\mathfrak{a}$. Since $\mathfrak{q}^c$ contains $\mathfrak{a}$ and is contained in $\mathfrak{p}$, minimality forces $\mathfrak{q}^c = \mathfrak{p}$, and hence $\mathfrak{q} = \mathfrak{p} A_\mathfrak{p}$. Since $\mathfrak{p} A_\mathfrak{p}$ is the unique prime containing $\mathfrak{a} A_\mathfrak{p}$, we have $\sqrt{\mathfrak{a} A_\mathfrak{p}} = \mathfrak{p} A_\mathfrak{p}$. In a Noetherian local ring $(A_\mathfrak{p}, \mathfrak{p} A_\mathfrak{p})$, an ideal whose radical is the maximal ideal is $\mathfrak{m}$-primary — this is part of the [Characterisation of $\mathfrak{m}$-Primary Ideals](/theorems/2900). [/guided] [/step] [step:Bound $\delta(A_\mathfrak{p}) \le r$ and apply the Dimension Theorem] Since $\mathfrak{a} A_\mathfrak{p}$ is a $\mathfrak{p} A_\mathfrak{p}$-primary ideal generated by $r$ elements $x_1/1, \ldots, x_r/1$, the $\delta$-invariant of $A_\mathfrak{p}$ satisfies: \begin{align*} \delta(A_\mathfrak{p}) \le r. \end{align*} This is because $\delta(A_\mathfrak{p})$ is defined as the infimum of the number of generators of $\mathfrak{p} A_\mathfrak{p}$-primary ideals, and we have exhibited a $\mathfrak{p} A_\mathfrak{p}$-primary ideal with $r$ generators. By the [Dimension Theorem](/theorems/860) applied to the Noetherian local ring $(A_\mathfrak{p}, \mathfrak{p} A_\mathfrak{p})$: \begin{align*} \operatorname{ht}(\mathfrak{p}) = \dim(A_\mathfrak{p}) = \delta(A_\mathfrak{p}) \le r. \end{align*} [guided] The $\delta$-invariant of a Noetherian local ring $(B, \mathfrak{m})$ is defined as: \begin{align*} \delta(B) = \min \{ n \ge 0 : \text{there exists an } \mathfrak{m}\text{-primary ideal of } B \text{ generated by } n \text{ elements} \}. \end{align*} We have produced a specific $\mathfrak{p} A_\mathfrak{p}$-primary ideal, namely $\mathfrak{a} A_\mathfrak{p}$, that is generated by $r$ elements. Therefore $\delta(A_\mathfrak{p}) \le r$ (the infimum over all such ideals is at most $r$). The [Dimension Theorem](/theorems/860) for a Noetherian local ring $(B, \mathfrak{m})$ states $\delta(B) = d(G_\mathfrak{m}(B)) = \dim(B)$. In particular, $\dim(A_\mathfrak{p}) = \delta(A_\mathfrak{p})$. Combining: \begin{align*} \operatorname{ht}(\mathfrak{p}) = \dim(A_\mathfrak{p}) = \delta(A_\mathfrak{p}) \le r. \end{align*} This completes the proof: every minimal prime $\mathfrak{p}$ over $\mathfrak{a} = (x_1, \ldots, x_r)$ has height at most $r$. [/guided] [/step]