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.
Best Answer
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.
EDIT. The sketch proof that you wrote is incorrect. As I think more about it, using the same exact sequence could give you a proof that one might expect, with a little more work.
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 $\ell(M/(y_1,...,y_i)M) < \infty$).