Krull dimension of polynomial rings over noetherian rings

Question

I'm trying to prove the following theorem involving Krull dimension:

Theorem If $A$ is a noetherian ring then $$\dim(A[x_1,x_2, \dots , x_n]) = \dim(A) + n$$ where $\dim$ stands for the Krull dimension of the rings. Thus, $\dim(K[x_1,x_2, \dots , x_n]) = n$ for any field $K$.

I know how to prove it by induction on $n$, but only if I assume the following facts:

1. If $A$ is noetherian then $A[x_1,x_2, \dots , x_n]$ is noetherian.
2. If $K$ is a field then $\dim(K)=0$.
3. If $A$ is noetherian then $\dim(A[x]) = \dim(A) + 1$.

Fact 1 follows from Hilbert's basis theorem. Fact 2 is trivial.

As to Fact 3, I was able to show that $\dim(A[x]) \geq \dim(A) + 1$, but couldn't prove the reverse inequality.

I have two questions:

1. How would a proof for the missing inequality look like? Is there an easy version assuming $A$ to be a PID?

2. Are there counterexamples to the theorem if we relax the condition of $A$ being noetherian to $A$ being finite dimensional?

Thanks in advance!

Answer 1

Here is an outline of a proof, based on the last exercise in Atiyah-Macdonald:

Take a prime ideal $\mathfrak{p}\subset A$ of maximal height $m$. It is sufficient to show that $\mathfrak{p}[X]$ has height $m$ in $A[X]$, because $A[X]/\mathfrak{p}[X] \equiv (A/\mathfrak{p})[X]$ has dimension $1$.

The first step is to find an ideal $\mathfrak{a}=(a_1,\ldots , a_m) \subset \mathfrak{p}$ such that $\mathfrak{p}$ is minimal over $\mathfrak{a}$.* The next step is to show that $\mathfrak{p}[X]$ is minimal over $\mathfrak{a}[X]$.** This tells us that $\mathfrak{p}$ has height at most $m$,*** and it is easy to show that it has height at least $m$.

* and *** seem to require something like the Hauptidealsatz, while ** may require primary decomposition.

Note that this proof is quite straightforward when $A$ is a PID.

There are known counterexamples where the dimension of $A[X]$ can be as large as $2m+1$. See my example here.

Answer 2

Let me recall your questions:

1. Is there an easy proof for $A$ a PID? Yes. Let $M$ be a maximal ideal in $A[X]$, and suppose the height of $M$ is $>2$. Then use that there is no chain of three primes in $A[X]$ lying over the same prime ideal of $A$ to get $(p)=M\cap A$, $p\ne0$ prime element. It follows that $pA[X]\subsetneq M$ and the height of $pA[X]$ is one, a contradiction.

2. The formula holds for rings which are not necessarily noetherian but have finite Krull dimension? Of course not! There are examples of rings $A$ of dimension $n$ with $\dim A[X]$ any number in the set $\{n+1,\dots,2n+1\}$.