[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 \(\mathbb{A}^k.\))
• . . .