Krull dimension of Nagata rings I want any help concerning proof of the following theorem of Nagata:
Let $S$ be the set of all $f\in R[x]$ with the property that the coefficients of $f$ generate the unit ideal. Then
(a) $S$ is a multiplicative set consisting entirely of non-zero-divisors;
(b) $\dim S^{-1}R[x]=\dim R$, where $\dim$ stands for the Krull dimension of $R$.
Thanks in advance!
 A: Denote by $c(f)$ the ideal generated by all the coefficients of $f\in R[X]$.
(a) Hint. $c(f)=R$ iff $\bar f\ne\bar 0$ in $(R/\mathfrak m)[X]$ for any maximal ideal $\mathfrak m$ of $R$.
(b) (Here you missed the condition $R$ noetherian, otherwise the result can be wrong.)
Clearly $\dim S^{-1}R[X]\ge\dim R$.
Now let $P$ be a prime ideal of $R[X]$ maximal with the property $P\cap S=\emptyset$. It's obvious that $S=R[X]\setminus\bigcup_{\mathfrak m\in\operatorname{Max}R}\mathfrak m[X]$. From $P\cap S=\emptyset$ we get $P\subseteq\bigcup_{\mathfrak m\in\operatorname{Max}R}\mathfrak m[X]$. Now we claim that there is $\mathfrak n\in\operatorname{Max}R$ with the property $P\subseteq\mathfrak n[X]$. In order to prove this let's consider $\mathfrak a$ the set of all elements of $R$ which appear as a coefficient of a polynomial from $P$. This is an ideal of $R$, and $\mathfrak a\ne R$. Then there is $\mathfrak n\in\operatorname{Max}R$ with the property $\mathfrak a\subseteq\mathfrak n$, and therefore $P\subseteq\mathfrak n[X]$. In particular, $\operatorname{ht} P\le\operatorname{ht}\mathfrak n[X]=\operatorname{ht}\mathfrak n$, and we are done.
