For a commutative ring $S$ of finite Krull dimension $d$, we have $1+d\leq \dim(S[X])\leq 2d+1$. One proof of this uses the fact that if $Q\_1\subset Q\_2\subset Q\_3$ is a chain of prime ideals of $S[X]$, then $Q\_1\cap S\subset Q\_3\cap S$.


Now let $R$ be a commutative ring of infinite dimension. Let $\mathcal{C}$ be a chain of prime ideals in $R[X]$ of arbitrary cardinality. Consider $\mathcal{C\_R}=\{Q\cap R: Q\in \mathcal{C}\}$, a chain of prime ideals in $R$. It follows from the fact in the previous paragraph that $\mathcal{C\_R}$ is countably infinite if and only if $\mathcal{C}$ is countably infinite. Is it true in general that $\mathcal{C}$ and $\mathcal{C\_R}$ have the same cardinality?



Arbitrary chains of prime ideals in $R[X]$
From the fact $Q\_1 \cap R \subsetneq Q\_3 \cap R$, we see that the natural map from $\mathcal{C}$ onto $\mathcal{C}\_R$ is at most 2-to-1. If $\mathcal{C}$ is infinite, this implies that it has the same cardinality as its image $\mathcal{C}\_R$. Am I missing something?


