15
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ I like the exposition in math.iitb.ac.in/atm/caag1/ghorpade.pdf, Proposition 4.7 (on page 40). $\endgroup$ Commented Feb 5, 2016 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
    Commented Feb 14, 2017 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. $\endgroup$ Commented Feb 14, 2017 at 11:36

2 Answers 2

12
$\begingroup$

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.

$\endgroup$
4
  • 7
    $\begingroup$ @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
    Commented Jun 8, 2017 at 4:01
  • $\begingroup$ You mean $> \dim A + 1$. $\endgroup$
    – Cloudscape
    Commented Jun 23, 2023 at 7:03
  • $\begingroup$ Just a comment, step ** does not require primary decomposition, it is direct. If $(a_1, \ldots, a_m) \subseteq Q \subseteq P[x]$, then intersecting with $A$ gives $(a_1, \ldots, a_m) \subseteq Q \cap A \subseteq P$, hence $Q \cap A = P$ by minimality, hence $P[x] \subseteq Q$ (and so they are equal). $\endgroup$ Commented Mar 19 at 4:00
  • 1
    $\begingroup$ @rmdmc89 you're right, there is an extra step. Specifically, if $Q \subsetneq Q'$ in $A[x]$ and $Q \cap A = Q' \cap A = P$, then $Q = P[x]$. Now a maximal chain in $A[x]$ must have such a pair. Since the chain is maximal, $\dim A = \mathrm{ht}(Q) + \dim A[x]/Q$, which by the other statements is $\mathrm{ht}(P) + \dim (A/P)[x]$, which by induction is $\leq \mathrm{ht}(P) + \dim A/P + 1 \leq \dim A + 1$. $\endgroup$ Commented Mar 21 at 18:50
12
$\begingroup$

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

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .