Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
I asked a similar question, though with $A$ a UFD, here on MO; perhaps it will be helpful. – Zev Chonoles Nov 21 '11 at 11:22
I fear that it is quite different if $A$ is a GCD domain, but not a Bézout domain. However thanks for the link! – Andrea Nov 21 '11 at 11:31
Point 3 follows from the facts that the polynomial ring over a normal domain is normal, and that normality ist stable under localization. – Hagen Knaf Nov 21 '11 at 11:49
Right! I was careless. – Andrea Nov 21 '11 at 12:01
up vote 2 down vote accepted

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 :).

share|cite|improve this answer
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 $\dim A=\dim A(x)$. – Hagen Knaf Nov 21 '11 at 16:05
Dear @Hagen, our proofs go the same way. – user18119 Nov 21 '11 at 16:11
@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 $2$. – Andrea Nov 21 '11 at 16:22
@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 $x=0$. – user18119 Nov 21 '11 at 17:03
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. – user18119 Nov 21 '11 at 17:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.