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?


  • $\begingroup$ "What is a proof for the missing inequality?" Is there any textbook in commutative algebra which doesn't prove this? $\endgroup$ – user26857 May 4 '15 at 22:30
  • $\begingroup$ I like the exposition in math.iitb.ac.in/atm/caag1/ghorpade.pdf, Proposition 4.7 (on page 40). $\endgroup$ – Ingo Blechschmidt Feb 5 '16 at 14:23
  • $\begingroup$ @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... $\endgroup$ – Watson Feb 14 '17 at 10:58
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.

  • 3
    $\begingroup$ @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 $\geq\dim A+1$ $\endgroup$ – rmdmc89 Jun 8 '17 at 4:01

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.