Prime ideals of the ring of rational functions Let $A$ be a commutative ring with identity. If $f = a_0 + a_1 x + \cdots + a_n x^n \in A[x]$ is a polynomial, define $c(f) = A a_0 + A a_1 + \cdots + A a_n$ the ideal of $A$ generated by the coefficients of $f$. Consider $S$ the subset of $A[x]$ made up of primitive polynomials, i.e. polynomials $f \in A[x]$ such that $c(f) = A$. It is not difficult to prove that $S$ is a multiplicative subset of $A[x]$. Consider the ring
$$
A(x) = S^{-1} (A[x]).
$$
It is easy to show that $S$ does not contain zero-divisors of $A[x]$, hence $A \subseteq A[x] \subseteq A(x)$. If $I$ is an ideal of $A$ then $(I \cdot A(x)) \cap A = I$. If $\mathfrak{p}$ is a prime ideal of $A$ then $\mathfrak{p} \cdot A(x)$ is a prime ideal of $A(x)$. The map $\phi \colon \mathrm{Spec} A(x) \to \mathrm{Spec} A$ defined by $\phi(P) = P \cap A$ is surjective, because a right-inverse is $\mathfrak{p} \mapsto \mathfrak{p} \cdot A(x)$. It is clear that $\dim A \leq \dim A(x) \leq \dim A[x]$.


*

*If $M$ is a maximal ideal of $A(x)$, then does there exist a maximal ideal $\mathfrak{m}$ of $A$ such that $M = \mathfrak{m} \cdot A(x)$?

*Is the map $\phi \colon \mathrm{Spec} A(x) \to \mathrm{Spec} A$ injective? If $A$ is noetherian, then $\dim A(x) = \dim A$?

*If $A$ is a normal domain, then is $A(x)$ a normal domain?
 A: The answer to (1) is yes. Let us first show 

If $I$ is an ideal of $A[x]$ such that $I\cap S=\emptyset$, then $I$ is contained in $\mathfrak m A[x]$ for some maximal ideal $\mathfrak m$ of $A$. 

Proof. Note that 
$$I\subseteq \sum_{f\in I} c(f)A[x].$$
So if $I$ is not contained in any $\mathfrak mA[x]$, then $A=\sum_{1\le i\le n} c(f_i)$ for some $f_i\in I$. Fix an $m$ big enough and write
$$f_i(x)=a_{i0}+a_{i1}x+...+ a_{im}x^m$$
($a_{im}$ could be zero) and an identity
$$1=\sum_{i\le n, j\le m} \alpha_{ij}a_{ij}, \quad \alpha_{ij}\in A.$$
Consider the polynomial
$$f=\sum_{i,j}\alpha_{ij}f_i(x)x^{m-j} \in I.$$
The term of degree $m$ in $f$ has coefficient equal to $1$. So $f\in S$. Contradiction. 
As any $\mathfrak m A[x]$ is prime and has empty intersection with $S$, the above result implies immediately (1). 
(2) The map $\phi$ is clearly surjective (consider $\mathfrak qA[x]$ for any $\mathfrak q\in\mathrm{Spec} A$) but is not injective in general. Consider $A=k[t,s]$ over a field $k$. Then $f(x):=t+sx$ generates a prime ideal $\mathfrak p$ of $A[x]$ which doesn't meet $S$ and we have $\mathfrak pS^{-1}A[x]\cap A=\{0\}$. So the generic fiber of $\phi$ has at least two points (in fact infinitely many points).  
Note however that $\phi$ is always injective over any maximal ideal $\mathfrak m$ because $\phi^{-1}(\mathfrak m)=\mathfrak mS^{-1}A[x]$. 
The argument for the surjectivity of $\phi$ also shows that $\dim S^{-1}A[x]\ge \dim A$. We also have $\dim S^{-1}A[x]\le \dim A[x]=\dim A +1$ when $A$ is noetherian. I don't know whether the equality is possible.
Update: We have $\dim S^{-1}A[x]=\dim A$ when $A$ is noetherian: let $\mathfrak p_0\subset ... \subset \mathfrak p_n$ be a chain of prime ideals of $A[x]$ contained in $A[x]\setminus S$. By (1), we have $\mathfrak p_n\subset \mathfrak mA[x]$ for some maximal ideal $\mathfrak m\subset A$. Then
$$ \mathfrak p_0\subset ... \subset \mathfrak p_n\subset \mathfrak mA[x]+xA[x]$$
is a chain of prime ideals of $A[x]$. Thus $\dim S^{-1}A[x]\le \dim A[x] -1=\dim A$. Your homework is not so easy :). 
