Converse to Krull's height theorem in localised form

By [L1] and [L2], the local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ and dimension $n$. Applying [L3] to the proper ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ produces elements $u_1,\dots ,u_n\in \mathfrak{p}{R}_{\mathfrak{p}}$ such that $J=(u_1,\dots ,u_n)$ has height $n$. Writing each $u_i=a_i\mathrm{/}{s}_i$ with $a_i\in \mathfrak{p}$ and $s_i\mathrm{\notin }\mathfrak{p}$, and replacing $u_i$ by the associate $s_i{u}_i=a_i\mathrm{/}1$, we may assume $u_i=x_i\mathrm{/}1$ with $x_i\in \mathfrak{p}$.

Let $\mathfrak{q}$ be a prime ideal minimal over $J=(x_1\mathrm{/}1,\dots ,x_n\mathrm{/}1)$. Because $J$ has height $n$, the prime $\mathfrak{q}$ has height $n$. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ also contains $J$, so $\mathfrak{q}\subseteq \mathfrak{p}{R}_{\mathfrak{p}}$. If the inclusion were strict, a strict chain of length $n$ ending at $\mathfrak{q}$ would extend by one more step to a chain ending at $\mathfrak{p}{R}_{\mathfrak{p}}$, contradicting $\dim (R_{\mathfrak{p}})=n$. Therefore $\mathfrak{q}=\mathfrak{p}{R}_{\mathfrak{p}}$, so the maximal ideal is minimal over $J$.

Therefore $\mathfrak{p}$ becomes minimal over an ideal generated by $n$ elements after localizing at $\mathfrak{p}$.