Let $(I,\leq)$ be a poset. Recall that the Krull dimension of $I$ is defined as follows:
— $K.dim(I)=-1$ if and only if $I=\{0\}$;
— if $\alpha$ is an ordinal and we already defined what it means to be a poset with Krull dimension $\beta$ for any ordinal $\beta<\alpha$, we say that $K.dim(I)=\alpha$ if and only if $K.dim(I)\neq \beta$ for all $\beta<\alpha$ and, for every descending chain
$$x_1\geq x_2\geq x_3 \geq \ldots \geq x_n\geq \dots$$
of elements of $I$, there exists $\bar n\in \mathbb N_+$ such that $K.dim([x_n,x_{n+1}])=\beta_n$ for all $n\geq \bar n$ and $\beta_n$ an ordinal $<\alpha$;
(here the notation $[a,b]$ is used for the segment between $a$ and $b$, that is, the subset of $I$ of all elements $\geq a$ and $\leq b$)
The above definition is mostly applied to posets which are indeed lattices. Notice that the lattices with $0$ Krull dimension are precisely the Artinian lattices, it is known (and not difficult to prove) that Noetherian lattices always have Krull dimension (this in fact generalizes to the following statement: "a lattice has Krull dimension if and only if its dual lattice has Krull dimension").
Let $A\subseteq \mathbb K^n$ be an affine algebraic variety over an algebraically closed field $\mathbb K$. It is well-known that, when endowed with its Zariski topology, $A$ is a Noetherian topological space, that is, the lattice of open subsets is Artinian, equivalently, the lattice of closed subsets is Noetherian. In particular, both these lattices have Krull dimension. There is a third natural lattice to consider, that is, the lattice $\mathcal C(A)$ of constructible subsets of $A$, which is the smallest lattice containing open and closed subsets of $A$ (that is, $\mathcal C(A)$ is obtained taking finite unions of intersections of an open and a closed subset).
Now, here is my question:
Can we say that $\mathcal C(A)$ has Krull dimension? Is this dimension equal to the dimension of the lattice of closed subsets?
Best Answer
For an arbitrary noetherian topological space $X$, I think the lattice $\mathcal{U}(X)$ of open subsets and the lattice $\mathcal{C}(X)$ of constructible subsets (the boolean algebra generated by open subsets) have the same Krull dimension (I use the standard ordering, to avoid confusion (see Ramiro's comment), so the lattice of closed subsets has Krull dimension zero by definition). I'll show they're both equal to a third number, based on the (finer) notion of ordinal length.
Let $\alpha\ge 1$ be an ordinal. Define $\log_\omega(\alpha)$ as the unique ordinal $\kappa$ such that $\omega^\kappa\le\alpha<\omega^{\kappa+1}$ (we agree that $\log_\omega(0)=-1)$.
Let $X$ be a noetherian topological space. Define inductively its length $\ell(X)=\sup(\ell(Y)+1)$ where $Y$ ranges over closed subsets of $X$ distinct of $X$ ($\sup\emptyset=\emptyset$). This is a standard inductive definition (take $Y$ minimal such that $\ell(Y)$ is not defined to get a contradiction). Note that $\ell(X)=0$ iff $X$ is empty.
Let's proceed to the proof.
Notation: We say that an ordinal $\alpha\ge 1$ is irreducible if $\alpha=\omega^\kappa$ for some $\kappa$ (necessarily equal to $\log_\omega(\alpha)$), or equivalently if $\beta,\gamma<\alpha$ implies $\beta+\gamma<\alpha$.
Also $\mathcal{C}(X;A,B)$, for $A,B\subset X$, is the interval between $A$ and $B$, i.e. those $Y\in\mathcal{C}(X)$ such that $A\subset Y\subset B$. Similarly $\mathcal{U}(X;A,B)$ is defined.
The proposition follows from the two following claims (1) (2).
We need to know the following facts on the length function (obtained by an easy induction): if $Y$ is a subset of $X$ with the induced topology, then $\ell(Y)\le\ell(X)$. Also if $F$ is a closed subset of $X$, then $\ell(X)\ge\ell(F)+\ell(X\smallsetminus F)$.
We prove (1) this by induction on $\ell(X)$.
Let $(C_n)$ be a descending chain of constructible subsets of $X$ and let us prove that $\mathcal{C}(X;C_{n+1},C_n)$ has Krull dimension $<\kappa$. If $C_n$ is eventually empty, $\mathcal{C}(X;C_{n+1},C_n)$ is eventually a singleton, thus of Krull dimension $-1<\kappa$. Now assume the contrary.
If for some $n$, $C_n$ is not dense, we can work inside its closure and argue by induction to get the desired conclusion. So we assume that each $C_n$ is dense (and not empty). It follows that the interior $U_n$ of $C_n$ is dense as well (in any noetherian topological space, a dense constructible subset has dense interior). In particular, $U_n$ is nonempty as well.
Then $(\ell(U_n))$ is a descending chain of ordinals, hence stabilizes, say for $n\ge n_0$. Set $V=U_{n_0}$. Then for $n\ge n_0$ we have $\ell(V)\ge\ell(V\smallsetminus U_n)+\ell(U_n)=\ell(V\smallsetminus U_n)+\ell(V)$. An easy consequence is that $\ell(V\smallsetminus U_n)<\omega^\kappa$ (using that every ordinal $<\omega^{\kappa+1}$ can be written as $\omega^\kappa\cdot n+\beta$ for a unique integer $n\ge 0$ and ordinal $\beta<\omega^\kappa$, particular case of the "Cantor form"). By induction, it follows that $V\smallsetminus U_n$ has Krull dimension $<\kappa$ for all $n\ge n_0$. Hence $\mathcal{C}(X,U_{n},V)$ has Krull dimension $<\kappa$ for all $n\ge n_0$.
Let $F$ be the complement of $V$. Then $\ell(F)<\ell(X)$, because $V\neq\emptyset$, so $F$ has Krull dimension at most $\kappa$. Hence there exists $n_1\ge n_0$ such that $\mathcal{C}(F;C_{n+1}\cap F,C_n\cap F)$ has Krull dimension $<\kappa$ for all $n\ge n_1$. The lattice $\mathcal{C}(X;C_{n+1},C_n)$ naturally embeds into the product of the lattices $\mathcal{C}(F;C_{n+1}\cap F,C_n\cap F)$ and $\mathcal{C}(V;C_{n+1}\cap V,C_n\cap V)$, and the latter is contained in $\mathcal{C}(V;U_{n+1},V)$. Since a product of lattices of Krull dimension $<\kappa$ also has Krull dimension $\kappa$, we deduce that $\mathcal{C}(X;C_{n+1},C_n)$ has Krull dimension $<\kappa$ for all $n\ge n_1$. This proves that $\mathcal{C}(X)$ has Krull dimension $\le\kappa$.
We'll use the immediate fact that if $X$ is a noetherian space, then for every ordinal $\gamma\le\ell(X)$ there exists a closed subset $Y$ of $X$ such that $\ell(Y)=\gamma$.
We also use that if $\kappa\ge 1$ and $Y$ is a closed subset of $X$ then $\ell(X)\le \ell(Y)\oplus\ell(X\smallsetminus Y)$, where $\oplus$ is the commutative sum of ordinals: $\alpha\oplus\beta=\max\{\sup_{\gamma<\alpha}((\gamma\oplus\beta)+1),\sup_{\delta<\alpha}((\alpha\oplus\delta)+1))\}$. In particular if $\ell(Y)$ and $\ell(X\smallsetminus Y)$ are both $<\omega^\kappa$ (for $\kappa\ge 1$) then so is $\ell(X)$.
To prove the claim (2), if $\kappa=0$ there is nothing to prove, so suppose $\kappa\ge 1$.
Suppose by contradiction the Krull dimension of $\mathcal{U}(X)$ is an ordinal $\beta<\kappa$. To get a contradiction, we have to construct a descending sequence $(U_n)$ of open subsets such that for each $n$, the Krull dimension of $\mathcal{U}(X;U_{n+1},U_n)$ is $\ge\beta$. Let $Y\neq X$ be a closed subset of $X$ with $\ell(Y)=\omega^\beta$. Define $U_1$ as the complement of $Y$. Since $\ell(Y)=\omega^\beta<\omega^\kappa$, it follows that $\ell(U_1)\ge\omega^\kappa$.
So we can continue by induction (on the integers) to obtain a descending sequence $(U_n)$ of open subsets, so that for each $n$ we have $\ell(U_n\smallsetminus U_{n+1})=\omega^\beta$. By induction (on ordinals), the Krull dimension $\mathcal{U}(U_n\smallsetminus U_{n+1})$ is $\ge\beta$. The latter lattice can be identified with the lattice $\mathcal{U}(X,U_{n+1},U_n)$. Hence we have a contradiction and it follows that the Krull dimension of $\mathcal{U}(X)$ is $\ge\kappa$.