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,
• $\varphi \in {\mathcal{L}}_x(M)$ a formula with parameters from $M$ and free variables ranging in $x,$ and
• $\alpha$ an ordinal

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

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

We define $\text{RM}(\varphi ):=\alpha$ if $\text{RM}(\varphi )\ge \alpha$ but $\text{RM}(\varphi )\ngeqq \alpha +1.$ 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 $\varphi .$ One can verify that $\text{RM}(\psi )=0$ when the solution set of $\psi$ in $M$ is finite, so $\text{RM}(\varphi )=0.$
• The affine line ${\mathbb{A}}^1(K)=K$ has dimension 1, and $\text{RM}(K)\ge 1.$ 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 ${\mathbb{A}}^k.$)
• . . .