# Krull dimension of $\mathbb{Z}[x_1,\dots,x_n]$

I'm trying to prove that the Krull dimension of $\mathbb{Z}[x_1,\dots, x_n]$ is $n+1$. I know there is a result that says $$\dim(A[x_1,\dots, x_n])=n+\dim(A),$$ when $A$ is a Noetherian ring, but I was outlined a proof of this by another method that I don't quite understand.

The idea is to use Hilbert polynomials and that if $F$ is a field, then $F[x_1,\dots x_n]$ has Krull dimension $n$.

It goes as follows:

Let $\mathfrak m$ be a maximal ideal of $\mathbb{Z}[x_1,\dots, x_n]$, and $p$ a prime number in $\mathfrak m$ (you can get this prime by looking at $\mathfrak m \cap \mathbb{Z}$ or at least that's what I think). Then consider the short exact sequence $$0 \longrightarrow \mathbb{Z}[x_1,\dots, x_n] \overset{p\cdot}\longrightarrow \mathbb{Z}[x_1,\dots, x_n] \longrightarrow \mathbb{F}_p[x_1,\dots, x_n] \longrightarrow 0.$$

(Here by $p\cdot$ I mean the map multiplication by $p$).

Then he does something I don't quite understand, he localizes this sequence at $\mathfrak m$ to get $$0 \longrightarrow \mathbb{Z}[x_1,\dots, x_n]_{\mathfrak m} \overset{p\cdot}\longrightarrow \mathbb{Z}[x_1,\dots, x_n]_{\mathfrak m} \longrightarrow \mathbb{F}_p[x_1,\dots, x_n]_{\mathfrak m} \longrightarrow 0$$ and finally uses Hilbert polynomials to get $$P(\mathbb{Z}[x_1,\dots, x_n]_{\mathfrak m},t)=P(\mathbb{Z}[x_1,\dots, x_n]_{\mathfrak m},t) \cdot t + P(\mathbb{F}_p[x_1,\dots, x_n]_{\mathfrak m},t).$$ Now here I don't understand why on the RHS $$P(\mathbb{Z}[x_1,\dots, x_n]_{\mathfrak m},t)$$ is multiplied by $t$. But from this I can see how the result follows at once.

Is this proof correct? And if so why do you localize at $\mathfrak m$ and why the factor $t$ on the RHS?

Thank you.

I think it is an incorrect proof. The reason why one localized at $m$ is to have Noetherian local rings (remember that the degree of Hilbert-Samuel polynomial = dimension holds for local Noetherian rings). The thing that goes wrong in the proof (based on what you wrote) is: in the last equality, the same Hilbert-Samuel $P(\mathbb{Z}[x_1,...,x_n]_m,t)$ appears too times and the later has strictly bigger degree, obviously impossible.
To simplify the notation, let us denote $A = \mathbb{Z}[x_1,...,x_n]_m$ and $B = \mathbb{F}_p[x_1, ..., x_n]_m$ and $m = mA$ (the last one is an abuse of notation meaning the unique maximal of $A$). Let us denote the Hilbert Samuel polynomial of $A$ and $B$ as $P$ and $Q$ respectively. You begin with your exact sequence $$0 \rightarrow A \rightarrow^{p} A \rightarrow B \rightarrow 0.$$ Tensor this sequence with $m^n/m^{n+1}$ and using Artin-Rees Lemma, you get an inequality $$0 \leq Q(n) \leq P(n) - P(n-c),$$ where $c \in \mathbb{Z}_+$ is the power that appears in Artin-Rees Lemma (there's a little detail to work out here and you can look it up in Atiyah-McDonald for instance). The above inequality tells you that $\deg Q \leq \deg P - 1$. Since $\deg Q = \dim B = n$ (the dimension of a polynomial ring over a field equals its number of variables), it remains to show that $\deg Q \geq \deg P - 1$. This results simply because $B \cong A/pA$ and $\dim A/pA \geq \dim A - 1$ (using the fact that $\dim M$ equals the minimal length $i$ of a sequence $y_1,...,y_i$ in $m$ such that $\ell(M/(y_1,...,y_i)M) < \infty$).
• In my comment, I did point out one particular equality in the proof is wrong, say $P(\mathbb{Z}[x_1,...,x_n]_m,t) = P(\mathbb{Z}[x_1,...,x_n]_m,t)t + P(\mathbb{F}[x_1,...,x_n]_m,t)$ and that would be enough to conclude that the particular proof is incorrect. If you want to be convinced that using the above exact sequence, will probably won't give you an expected proof, you need to dig into the proof that whenever you have an exact sequence $0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$, $dim N = \max\{dim M,dim P\}$ to get some feeling and intuition. – mr.bigproblem Apr 30 '13 at 21:00
• @ Kerami: I edited my answer. It turns out that my feeling about the exact sequence won't give a proof is wrong. In fact, there IS a proof using the same exact sequence, in my answer. I hope you are satisfied now. My last comment on the proof: the fact that proof works is mainly due to the fact that $\mathbb{Z}$ is a P.I.D. so that every time you take a maximal (prime) ideal of $\mathbb{Z}[x_1,...,x_n]$ intersect $\mathbb{Z}$, you always get a principal prime ideal in $\mathbb{Z}$. – mr.bigproblem Apr 30 '13 at 21:50
• @mr.bigproblem: I upvote the first part of your proof. But in the second part, at the end, you use the result that $\dim A/pA\ge \dim A-1$. As the inverse inequality trivially holds, you get the desired equality $\dim A=n+1$, and there is no need to use Hilbert-Samuel. I think you should improve the estimate about the contribution of $pA$. – user18119 May 1 '13 at 5:33