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 $\dim A \leq \dim A(x) \leq \dim A[x]$.

  1. If $M$ is a maximal ideal of $A(x)$, then does there exist a maximal ideal $\mathfrak{m}$ of $A$ such that $M = \mathfrak{m} \cdot A(x)$?
  2. Is the map $\phi \colon \mathrm{Spec} A(x) \to \mathrm{Spec} A$ injective? If $A$ is noetherian, then $\dim A(x) = \dim A$?
  3. If $A$ is a normal domain, then is $A(x)$ a normal domain?
  • $\begingroup$ I asked a similar question, though with $A$ a UFD, here on MO; perhaps it will be helpful. $\endgroup$ – Zev Chonoles Nov 21 '11 at 11:22
  • $\begingroup$ I fear that it is quite different if $A$ is a GCD domain, but not a Bézout domain. However thanks for the link! $\endgroup$ – Andrea Nov 21 '11 at 11:31
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    $\begingroup$ Point 3 follows from the facts that the polynomial ring over a normal domain is normal, and that normality ist stable under localization. $\endgroup$ – Hagen Knaf Nov 21 '11 at 11:49
  • $\begingroup$ Right! I was careless. $\endgroup$ – Andrea Nov 21 '11 at 12:01

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 :).

  • $\begingroup$ If $A$ is a local noetherian domain with maximal ideal $m$, then $S=A[x]\setminus m[x]$. Moreover any chain of primes in $A$ lifts to a chain of primes of $A[x]$ of the same length. Consequently $\dim A=\dim A(x)$. $\endgroup$ – Hagen Knaf Nov 21 '11 at 16:05
  • $\begingroup$ Dear @Hagen, our proofs go the same way. $\endgroup$ – user18119 Nov 21 '11 at 16:11
  • $\begingroup$ @QiL, the first assertion in your answer is illuminant. Thanks a lot! However, this is not homework; this is a generalization of mine to an exercise that I read some time ago and considered only the case when $A$ is a PID. This generalization comes from my attempt to find a finite extension of Dedekind domains $A \subseteq B$ such that $\mathrm{Frac}(B) / \mathrm{Frac}(B)$ is Galois and there exists a prime $\mathfrak{q}$ of $B$ such that $k(\mathfrak{q})/k(A \cap \mathfrak{q})$ is not separable; I think I can take $\mathbb{Z}(x^2) \subseteq \mathbb{Z}(x)$ and the prime generated by $2$. $\endgroup$ – Andrea Nov 21 '11 at 16:22
  • $\begingroup$ @Andrea There is a simpler example of $A\subseteq B$ as you want. Let $k$ be an imperfect field of characteristic $p$, let $a\in k\setminus k^p$. Let $A=k[x]$ and $B=k[x,y]$ with $y^p+xy+a=0$. This defines an Artin-Schreier cyclic extension on the generic fibers. Check that $B$ is a Dedekind domain and look at what happens above $x=0$. $\endgroup$ – user18119 Nov 21 '11 at 17:03
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    $\begingroup$ To have an example in characteristic zero: let $A$ be a DVR of characteristic 0 containing a $p$-th root of unit with imperfect residue field $k$ (of char. $p$), let $a\in A$ whose class in $k$ is not a $p$-th power and let $B=A[t]/(t^p-a)$. Then $A\subset B$ does the trick. $\endgroup$ – user18119 Nov 21 '11 at 17:07

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