A prime chain in R[x] has length at most one more than its contraction chain

Put $S=R\setminus \mathfrak{q}$, $B=R_{\mathfrak{q}}$, $\mathfrak{m}=\mathfrak{q}B$, and $P^{\mathrm{\prime }}=S^{-1}P\subset B[x]$. Every prime below $P$ contracts into $\mathfrak{q}$ and is disjoint from $S$, so localization preserves all chains below $P$ and $\mathrm{ht}(P)=\mathrm{ht}(P^{\mathrm{\prime }})$. Contraction from $B[x]$ also shows that $P^{\mathrm{\prime }}=\mathfrak{m}B[x]$ exactly when $P=\mathfrak{q}R[x]$. Likewise $\mathrm{ht}(\mathfrak{m})=r$. Under the assumed Choice hypothesis, [L1] supplies $a_1,\dots ,a_r\in \mathfrak{m}$ such that $\mathfrak{m}$ is minimal over $J=(a_1,\dots ,a_r)$ in $B$. Because $B$ is local with maximal ideal $\mathfrak{m}$, it is the only prime of $B$ containing $J$.

If $P^{\mathrm{\prime }}=\mathfrak{m}B[x]$, any prime $Q\subseteq P^{\mathrm{\prime }}$ containing $JB[x]$ contracts to a prime of $B$ containing $J$, hence to $\mathfrak{m}$ by step 1.1. Thus $Q$ contains $\mathfrak{m}B[x]=P^{\mathrm{\prime }}$ and equals it, proving that $P^{\mathrm{\prime }}$ is minimal over $JB[x]$. For $r\ge 1$, [L2] gives $\mathrm{ht}(P^{\mathrm{\prime }})\le r$; for $r=0$, $P^{\mathrm{\prime }}$ is a minimal prime of $B[x]$ because $\mathfrak{m}$ is the only prime of $B$, so its height is zero. Conversely a chain of length $r$ below $\mathfrak{q}$ extends coefficientwise to a chain of prime ideals below $\mathfrak{q}R[x]$, giving $\mathrm{ht}(P^{\mathrm{\prime }})\ge r$. Therefore this case has height exactly $r$.

If $P^{\mathrm{\prime }}\mathrm{\ne }\mathfrak{m}B[x]$, its image $P^{\mathrm{\prime }}\mathrm{/}\mathfrak{m}B[x]$ is a nonzero prime of the PID $(B\mathrm{/}\mathfrak{m})[x]$. By [L3], choose $f\in P^{\mathrm{\prime }}$ lifting a generator of that prime. Any prime $Q\subseteq P^{\mathrm{\prime }}$ containing $J$ and $f$ contracts to $\mathfrak{m}$ by step 1.1, hence contains $\mathfrak{m}B[x]$. Its image modulo $\mathfrak{m}B[x]$ contains the generator of $P^{\mathrm{\prime }}\mathrm{/}\mathfrak{m}B[x]$, and containment in $P^{\mathrm{\prime }}$ forces equality. So $Q=P^{\mathrm{\prime }}$ and $P^{\mathrm{\prime }}$ is minimal over $(J,f)$, which has $r+1\ge 1$ generators. By [L2], $\mathrm{ht}(P^{\mathrm{\prime }})\le r+1$. The extended coefficientwise chain below $\mathfrak{q}R[x]$, followed by the strict inclusion $\mathfrak{q}R[x]⊊P$, gives the reverse inequality. Hence $\mathrm{ht}(P)=r+1$.

For $R\mathrm{\ne }0$, consider any strict chain ending at a prime $P$ of $R[x]$. If $\mathrm{ht}(P\cap R)$ is finite, steps 2.1–2.2 bound its length by $\mathrm{ht}(P\cap R)+1\le \dim R+1$. If the contraction has infinite height, then $\dim R=\mathrm{\infty }$ and the claimed extended-integer bound is automatic.

Thus the exact relative height formula and the original chain bound both hold under the stated Choice hypothesis.