# Arbitrary chains of prime ideals in $R[X]$

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?

• Just to be sure: you don't assume noetherian, right? – YCor Mar 26 '15 at 14:50
• @YCor I don't assume noetherian, but I'm willing to consider noetherian as a special case if the conjecture is not true in general. – Andrew Chiriac Mar 26 '15 at 14:59
• It's that in the noetherian case the chains are well-ordered (for the reverse inclusion order) and then it's natural to consider the ordinal rather than the cardinal, to get a much more refined notion of Krull dimension. (Btw, are there examples in the non-noetherian case where $\dim(S[X])>\dim(S)+1$?) – YCor Mar 26 '15 at 15:05
• Yes, for every $n$ and $m$ with $1+n\leq m\leq 2n+1$ there is an integral domain $D$ such that $\dim(D)=n$ and $\dim(D[X])=m$. See the paper "A note on the dimension theory of rings (II)" by A. Seidenberg. – Andrew Chiriac Mar 26 '15 at 15:55

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?