110.21 Dimension in Noetherian Jacobson rings

Let $k$ be the algebraic closure of a finite field. Let $A = k[x, y]$ and $X = \mathop{\mathrm{Spec}}(A)$. Let $C = V(x)$ be the $y$-axis (this could be any other $1$-dimensional integral closed subscheme of $X$). Let $C_1, C_2, C_3, \ldots$ be an enumeration of the other integral closed subschemes of $X$ of dimension $1$. Let $p_1, p_2, p_3, \ldots$ be an enumeration of the closed points of $C$.

Claim: for every $n$ there exists an irreducible closed $Z_ n \subset X$ of dimension $1$ such that

$\{ p_ n\} = Z_ n \cap (C \cup C_1 \cup \ldots \cup C_ n)$

set theoretically. To do this set $Y = C \cup C_1 \cup C_2 \cup \ldots \cup C_ n$. This is a reduced affine algebraic scheme of dimension $1$ over $k$. It is enough to find $f \in k[x, y]$ with $V(f) \cap Y = \{ p_ n\}$ set theoretically because then we can take $Z_ n$ to be a suitable irreducible component of $V(f)$. Since the restriction map

$k[x, y] \longrightarrow \Gamma (Y, \mathcal{O}_ Y)$

is surjective, it suffices to find a regular function $g$ on $Y$ whose zero set is $\{ p_ n\}$ set theoretically. To see this is possible, we choose an effective Cartier divisor $D \subset Y$ whose support is $p_ n$ (this is possible by Varieties, Lemma 33.38.3). Thus it suffices to show that $\mathcal{O}_ X(ND) \cong \mathcal{O}_ X$ for some $N > 0$. But the Picard group of an affine $1$-dimensional algebraic scheme over the algebraic closure of a finite field is torsion (insert future reference here) and we conclude the claim is true.

Choose $Z_ n$ as above for all $n$. Since $k[x, y]$ is a UFD we may write $Z_ n = V(f_ n)$ for some irreducible element $f_ n \in A$. Let $S \subset k[x, y]$ be the multiplicative subset generated by $f_1, f_2, f_3, \ldots$. Consider the Noetherian ring $B = S^{-1}A$.

Obviously, the ring map $A \to B$ identifies local rings and induces an injection $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Moreover, looking at the curve $C_1$ we see that only the points of $C \cap C_1$ are removed when passing from $\mathop{\mathrm{Spec}}(A)$ to $\mathop{\mathrm{Spec}}(B)$. In particular, we see that $\mathop{\mathrm{Spec}}(B)$ has an infinite number of maximal ideals corresponding to maximal ideals of $A$. On the other hand, $xB$ is a maximal ideal because the spectrum of $B/xB$ consists of a unique prime ideal as we removed all the closed points of $C = V(x)$ (but not the generic point). Finally, for $i \geq 1$ consider the curve $C_ i$. Write $C_ i = V(g_ i)$ for $g_ i \in A$ irreducible. If $C_ i = Z_ n$ for some $n$, then $g_ iB$ is the unit ideal. If not, then all but finitely many of the closed points of $C_ i$ survive the passage from $A$ to $B$: namely, only the points of $(Z_1 \cup \ldots \cup Z_{i - 1} \cup C) \cap C_ i$ are removed from $C_ i$.

The structure of the prime spectrum of $B$ given above shows that $B$ is Jacobson by Algebra, Lemma 10.61.4. The maximal ideals are the maximal ideals of $A$ which are in $\mathop{\mathrm{Spec}}(B)$ (and there an inifinitude of these) together with the maximal ideal $xB$. Thus we see that we have local rings of dimensions $1$ and $2$.

Lemma 110.21.1. There exists a Jacobson, universally catenary, Noetherian domain $B$ with maximal ideals $\mathfrak m_1, \mathfrak m_2$ such that $\dim (B_{\mathfrak m_1}) = 1$ and $\dim (B_{\mathfrak m_2}) = 2$.

Proof. The construction of $B$ is given above. We just point out that $B$ is universally catenary by Algebra, Lemma 10.105.4 and Morphisms, Lemma 29.17.5. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).