# 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:

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.

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!

• "What is a proof for the missing inequality?" Is there any textbook in commutative algebra which doesn't prove this? – user26857 May 4 '15 at 22:30
• I like the exposition in math.iitb.ac.in/atm/caag1/ghorpade.pdf, Proposition 4.7 (on page 40). – Ingo Blechschmidt Feb 5 '16 at 14:23
• @IngoBlechschmidt : the link is broken. Maybe the corollary 3.14 here (by S. R. Ghorpade) is equivalent to the proposition 4.7 in your document... – Watson Feb 14 '17 at 10:58
• @Watson: Thanks for noticing. Luckily, the Internet Archive has a copy of the document. The notes you linked are quite nice, however it seems that they don't include a proof of the statement in question which is as detailed as in the older notes. – Ingo Blechschmidt Feb 14 '17 at 11:36

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.

• @Pipicito Updated with an example. – Slade May 5 '15 at 2:13
• @Slade, shouldn't we prove that every chain of primes in $A[x]$ has an element $p[x]$ where $p$ has maximal height? Otherwise, a priori we could have some different kind of chain with lenght possibly $\geq\dim A+1$ – rmdmc89 Jun 8 '17 at 4:01

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