TITLE: Krull dimension of polynomial rings over noetherian rings
QUESTION [13 upvotes]: I want to prove the following theorem concerning 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 could prove it by induction in the number of variables assuming these facts:

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

Fact 1 follows from Hilbert's basis theorem. Fact 2 is trivial.
For Fact 3 I have been able to prove the inequality $\dim(A[x]) \geq \dim(A) + 1$. I couldn't get the other inequality.
My questions are:

What is a proof for the missing inequality? Are there easy proofs if $A$ is assumed to be a PID?
Are there counterexamples to the theorem I'm trying to prove if we change the condition of $A$ being noetherian to $A$ being finite dimensional?

Thanks!

REPLY [11 votes]: Let me recall your questions: 

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.
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 $\{n+1,\dots,2n+1\}$.