Consider the polynomial ring $R[X]$ in one variable over a (not necessarily Noetherian) ring $R$. Prove the following inequality: $\operatorname{dim}R[X] \le 2\operatorname{dim}R+1$.
In Bosch's book there is a hint saying:
if $p_1 \subsetneq p_2 \subset R[X]$ are two (different) prime ideals in $R[X]$ restricting to the same prime ideal $p \subset R$, then $p_1 = pR[X]$.
Let $d:=\operatorname {dim}R$, and take an ascending chain of primes $p_1 \subsetneq p_2 \subsetneq\dots \subsetneq p_n\subset R[X]$. If, for some index $i$, $p_i\cap R=p_{i+1}\cap R=:p$, then the chain around $i$ becomes $$\dots\subsetneq pR[X]\subsetneq p_{i+1}\subsetneq p_{i+2}\subsetneq\dots$$ so that $p_{i+1}\cap R\neq p_{i+2}\cap R$. Indeed, in that case we would have $p_{i+1}=pR[X]=p_i$, against the hypothesis on the chain. Conclusion: the chain restricted to $R$ cannot have more than two consecutive equal primes.
Let $P_1 \subseteq P_2 \subseteq\dots \subseteq P_n\subset R$ be the restricted chain. From the previous point we can't cancel more than two consecutive ideals, and we can assume not to cancel $P_1$ (if it is equal to $P_2$, cancel $P_2$). Then if $n=2m+1$, we can't cancel more than $m$ ideals; in order to cancel $m+1$ ideals, we should necessarily cancel $P_1$ and $P_n$.
So if $n=2m$, the chain restricted to $R$ becomes of length at least $m$; if $n=2m+1$, the chain restricted to $R$ becomes of length at least $m+1$. In the latter case, using $n$ and $d$ one can write $\frac{n+1}2\le d$, and so $n\le 2d-1$; in the first case one obtains $n\le 2d$.
If I should guess where is the mistake, I would say that a chain of length $2m+1$ can become of length $m$ when restricted (because then we'd get the thesis with the argument in the last paragraph); but again, if for any ideal that we cancel there must be another ideal equal to it (that we don't cancel), how can be possible that we cancel $m+1$ primes from a chain of length $2m+1$, as $2(m+1)>2m+1$? Thanks for any clarify