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Let $f$ be in $\mathbb{Z}[x,y]$ and consider the the quotient ring $\mathbb{Z}[x,y] / \langle f \rangle$. The ring $\mathbb{Z}[x,y]$ has dimension 3, and the codimension of $\langle f \rangle$ is $\le 1$ by Krull's Ideal Theorem. Is it possible to deduce from this, or by some other method, that $\mathbb{Z}[x,y] / \langle f \rangle$ cannot have dimension 1?

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up vote 3 down vote accepted

OK. Every maximal ideal of $\mathbb{Z}[X,Y]$ has height 3. Once we know this, let $0\neq f\in \mathbb{Z}[X,Y]$, and pick a maximal ideal $\mathfrak{m}$ such that $f\in \mathfrak{m}$. Then the dimension of $\mathbb{Z}[X,Y]/(f)$ is $\geq$ the dimension of $(\mathbb{Z}[X,Y]/(f))_{\mathfrak{m}}$, so we come to the case of Noetherian local ring, and the later ring has dimension $\geq 3-1=2$.

OK, perhaps, I donot need to show every maximal ideal of $\mathbb{Z}[X,Y]$ has height 3. Let us find a maximal ideal $\mathfrak{m}$ with height 3 such that $f\in \mathfrak{m}$. First, find a prime number $p$ such that $f$ is not a unit in $\mathbb{Z}/p\mathbb{Z}[X,Y]$. Then find a maximal ideal of $\mathbb{F}_p[X,Y]$ such that $f$ is in it. This maximal ideal has height 2 which corresponds a maximal ideal of $\mathbb{Z}[X,Y]$ having height 3. We are done.

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Excellent, thank you. – Paul Slevin May 22 '12 at 11:29
I indeed said this in your previous question. Because for my poor English expression, I am sorry maybe I have not said too clear. You are welcome. – wxu May 22 '12 at 11:31
But what do you mean when you say $\mathbb{Z}[X,Y] / \langle f \rangle $ localised at $\mathfrak m$. this is not an ideal of that ring? Or do you mean the corresponding max ideal in the quotient? – Paul Slevin May 22 '12 at 11:35
I really donot know how to expain more. Localized at $\mathfrak{m}/(f)$ if you prefer, and this ring is isomorphic to $\mathbb{Z}[X,Y]_{\mathfrak{m}}/(f)$, taking localization and taking quotient commutes! – wxu May 22 '12 at 11:39
@Paul: I'm sorry I didn't get back to your questions sooner and glad to see that you reposted your final question. For the record, what wxu suggests with respect to a prime number is what I was hinting at (too obliquely, it now seems). – Pete L. Clark May 22 '12 at 18:42

Let me prove the following general result alluded to by wxu for $n=2$, for which I know no reference :

Theorem Every maximal ideal $\mathfrak m\subset \mathbb Z[X_1,...,X_n]$ has height $n+1$

Step 1
The quotient $\kappa(\mathfrak m)= \mathbb Z[X_1,...,X_n]/\mathfrak m \;$ is a field, finitely generated over $\mathbb Z$.
This implies that it has positive characteristic $p$ and is finite: $\mathbb F_p\subset \kappa(\mathfrak m)=\mathbb F_q$
(Bourbaki, Comm.Alg., Ch.5, §3, Thm.3, Cor.1)

Step 2
Since $p\in \mathfrak m$, the quotient morphism $\mathbb Z[X_1,...,X_n]\to \mathbb F_p[X_1,...,X_n]$ will send $\mathfrak m$ to the maximal ideal $\overline {\mathfrak m}\subset \mathbb F_p[X_1,...,X_n]$
However it is well known that every maximal ideal of $k[X_1,...,X_n]$ has height $n$ if $k$ is a field, so that $\overline {\mathfrak m} \;$ has height $n$. (Matsumura, Commutative algebra,14.H)

Step 3
Let $(0)\subsetneq \overline {\mathfrak p_1} \subsetneq ...\subsetneq \overline {\mathfrak p_n}=\overline {\mathfrak m}$ be a maximal chain of prime ideals in $\mathbb F_p[X_1,...,X_n]$ .
The chain of primes $(0)\subsetneq (p)\subsetneq\mathfrak p_1 \subsetneq ...\subsetneq \mathfrak p_n=\mathfrak m $ proves that $\operatorname {height}(\mathfrak m)\geq n+1$.
Since $\operatorname {dim }\: ( \mathbb Z[X_1,...,X_n])=n+1$, we must have $$\text {height}(\mathfrak m)= n+1\;$$ Done

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