Krull's height theorem

If $n=1$, [L1] gives $\mathrm{ht}(\mathfrak{p})\le 1$.

Assume $n\ge 2$ and that the theorem is known for $(n-1)$-generated ideals. Suppose for contradiction that $\mathrm{ht}(\mathfrak{p})\ge n+1$. Then there exists a strict prime chain $\mathfrak{p}={\mathfrak{p}}_d⊋{\mathfrak{p}}_{d-1}⊋\cdots ⊋{\mathfrak{p}}_0$ with $d\ge n+1\ge 3$. Under the assumed AC, [L2] replaces $x_1$ by a suitable element $b\in {\mathfrak{p}}_1$ of a same-length chain for which $\mathfrak{p}$ is minimal over $(b,x_2,\dots ,x_n)$.

By [L3], the quotient prime $\mathfrak{p}\mathrm{/}(b)$ is minimal over the $(n-1)$-generated ideal generated by the images of $x_2,\dots ,x_n$, and the above chain descends to a strict chain of length $d-1\ge n$ ending at $\mathfrak{p}\mathrm{/}(b)$ in $R\mathrm{/}(b)$. This contradicts the induction hypothesis for $(n-1)$ generators.

Therefore the assumption $\mathrm{ht}(\mathfrak{p})\ge n+1$ is impossible, and $\mathrm{ht}(\mathfrak{p})\le n$.