# Krull dimension of polynomial rings over noetherian rings

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?

• I like the exposition in math.iitb.ac.in/atm/caag1/ghorpade.pdf, Proposition 4.7 (on page 40). Commented Feb 5, 2016 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... Commented Feb 14, 2017 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. Commented Feb 14, 2017 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.

• @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 $\geq\dim A+1$ Commented Jun 8, 2017 at 4:01
• You mean $> \dim A + 1$. Commented Jun 23, 2023 at 7:03
• 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). Commented Mar 19 at 4:00
• @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 $\leq \mathrm{ht}(P) + \dim A/P + 1 \leq \dim A + 1$. Commented Mar 21 at 18:50

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