morley rank as dimension

February 02, 2026

[notes]

Let $\mathcal{L}$ be a first order language and let $T$ be a complete $\mathcal{L}$ theory. Given

• $M \models T$ a model of $T$,
• $x$ a variable context,
• $\phi \in \mathcal{L}_x(M)$ a formula with parameters from $M$ and free variables ranging in $x,$ and
• $\alpha$ an ordinal

we define the Morley rank $\text{RM}(\phi)$ of a formula $\phi$ by transfinite recursion on the condition $\text{RM}(\phi) \geq \alpha.$ The recursion satisfies:

• $\text{RM}(\phi) \geq 0$ if and only if $M \models (\exists x)\phi$
• $\text{RM}(\phi) \geq \alpha + 1$ if and only if there is some $N \succeq M$ (an elementary extension) and a sequence of formulas $(\psi_i)_{i \in \omega} \subseteq \mathcal{L}_x(N)^\omega$ such that for each $i,$ $N \models \psi_i \to \phi$ and $\text{RM}(\psi_i) \geq \alpha$ both hold and for $i \neq j$ we have that $N \models \psi_i \to \neg \psi_j.$
• $\text{RM}(\phi) \geq \lambda$ for a limit ordinal $\lambda$ if and only if $\text{RM}(\phi) \geq \alpha$ for all $\alpha < \lambda.$

We define $\text{RM}(\phi) := \alpha$ if $\text{RM}(\phi) \geq \alpha$ but $\text{RM}(\phi) \not\geq \alpha + 1.$ Every formula in a totally transcendental theory has ordinal-valued Morley rank. Inconsistent formulas are given Morley rank of $-1.$

When taking $T = \text{ACF}_p$ (the theory of algebraically closed fields for characteristic $p$), the Morley rank and Krull dimension agree (on constructible sets). Examples:

• Take finite $X \subset K^n.$ This has dimension 0, and can be specified by some formula $\phi.$ One can verify that $\text{RM}(\psi) = 0$ when the solution set of $\psi$ in $M$ is finite, so $\text{RM}(\phi) = 0.$
• The affine line $\mathbb{A}^1(K) = K$ has dimension 1, and $\text{RM}(K) \geq 1.$ However, it cannot be greater than 2 because every definable subset of $K$ is either finite or cofinite, and cofinite sets cannot be disjoint. (Similar arguments hold for $\mathbb{A}^k.$)
• . . .