Krull dimension of polynomial rings over noetherian rings 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!
 A: 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.
A: 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 $\{n+1,\dots,2n+1\}$.
