If $A$ is a commutative ring with unity, then a fact proved in most commutative algebra textbooks is:

$$\dim A + 1\leq\dim A[X] \leq 2\dim A + 1$$

Idea of proof: each prime of $A$ in a chain can arise from at most two prime ideals of $A[X]$.

The left equality holds when $A$ is a noetherian ring, but this paper of Seidenberg shows that, for rings of fixed dimension, every value in the interval is achievable.

There seems to be a shortage of elementary references on these counterexamples, however, so I thought it might be good to collect some specific examples of commutative rings $A$ with $\dim A[X] > \dim A$, or knowledge about simple cases.

In particular, it would be good to have some examples of rings $A$ with $\dim A = 1$ and $\dim A[X]=3$.

Here is an example I just worked out, inspired by Seidenberg's paper.

Let $k$ be a field, with $t$ transcendental over $k$, and let $A = \{f\in k(t)[[Y]] \mid f(0) \in k\}$. Then $A$ has unique nonzero prime ideal $P = \{f\in k(t)[[Y]] \mid f(0) = 0\}$, with $A/P \cong k$, so $\dim A = 1$.

To see that $\dim A[X]=3$, it is enough to exhibit a prime ideal lying between $(0)$ and $P[X]$. Consider the map $A[X] \to k(t)[[Y]]$ given by $p(X) \mapsto p(t)$. The image is an integral domain, so the kernel $K$ is prime. But $(YX-tY)\in K$, while $Y\notin K$, so $(0)\subsetneq K \subsetneq P[X]$.

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  • Possibly related: https://math.stackexchange.com/questions/1207818/, https://mathoverflow.net/questions/130687 – Watson Oct 27 '16 at 15:04

1 Answers1


Here is a classification of all integral domains $A$ with $\dim A=1$, $\dim A[X]=3$:

First, consider the map $\operatorname{Spec} A[X]\to\operatorname{Spec} A$ given by $\mathfrak{p}\to\mathfrak{p}\cap A$. The fiber over $P$ is isomorphic to $\operatorname{Spec} A_P / PA_P\otimes_A A[X] \cong (A_P / PA_P)[X]$, which is one-dimensional. It follows that a chain of three distinct primes in $A[X]$ cannot have the same intersection with $A$.

Suppose that there is a chain of four prime ideals in $A[X]$. By the previous paragraph, intersecting with $A$ must give us $(0)\subset (0) \subset P \subset P$ for some maximal $P\subset A$. It follows that the first three elements of the chain are $(0) \subsetneq \mathfrak{q} \subsetneq P[X]$, with $\mathfrak{q}\cap A = (0)$. The existence of such a $\mathfrak{q}$ is therefore equivalent to $\dim A[X] = 3$.

Then the map $A[X]\to K(A[X]/\mathfrak{q})$ restricts to an injection $A\to K(A[X]/\mathfrak{q})$, which induces an injection $K(A)\to K(A[X]/\mathfrak{q})$. So $\mathfrak{q}$ is the kernel of a map from $A[X]$ to a field extension of $K(A)$. Let $t$ be the image of $X$ under this map.

Since $\mathfrak{q}\neq (0)$, $t$ is algebraic over $K(A)$. Since $\mathfrak{q}\subset P[X]$, $t$ is transcendental over $K(A/P)$.

So all examples arise from a dimension $1$ integral domain $A$ with a maximal ideal $P$, such that there is some $t\in \overline{K(A)}$ that is transcendental over $K(A/P)$—i.e. every integral equation satisfied by $t$ has coefficients in $P$.

(Note: In the original example, $t$ satisfies the polynomial $p(T) = YT - tY$, so this is probably close to the simplest example of an element of $K(A)$ transcendental over $K(A/P)$.)

Conversely, such a $t$ gives rise to our desired prime ideal $\mathfrak{q}$.

Can the condition $t\in \overline{K(A)}$ be simplified to $t\in K(A)$? If so, the condition is much simpler: we must have some $t\in K(A)$ that can only be written as a fraction with both numerator and denominator in $P$.

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  • What is $K(-)$? Fraction field? Also, you say that "WLOG it is a domain", which seems to be reducing $A$ modulo minimal prime (is that right?). How do you reconstruct back all the possibly non-domain $A$'s? I see it can be done mechanically, but i there some "natural formulation"? – Pavel Čoupek May 05 '15 at 05:59
  • @PavelČoupek It's the fraction field. And I think that the most natural formulation is "$A$ has a minimal prime $P$ such that $A/P$ has the property stated." I kind of consider the question of finding all such $A$ to be a separate one, and in general I'm not sure whether this is tractable for non-noetherian rings. – Slade May 05 '15 at 16:55