{"id":418,"date":"2026-09-03T06:04:33","date_gmt":"2026-09-03T06:04:33","guid":{"rendered":"https:\/\/wp.math.berkeley.edu\/model-theory\/?p=418"},"modified":"2026-09-03T06:08:06","modified_gmt":"2026-09-03T06:08:06","slug":"an-introduction-to-non-minimality-degree","status":"publish","type":"post","link":"https:\/\/wp.math.berkeley.edu\/model-theory\/2026\/09\/03\/an-introduction-to-non-minimality-degree\/","title":{"rendered":"An introduction to non-minimality degree"},"content":{"rendered":"

The notion of the nonminimality degree of a stationary type was introduced by James Freitag and Rahim Moosa in the paper Bounding nonminimality and a conjecture of Borovik-Cherlin, J. Eur. Math. Soc. 27, 589 \u2013 613 (2025)<\/a>.\u00a0 \u00a0They were motivated by the problem of analyzing differential algebraic relations amongst solutions of differential equations and with R\u00e9mi Jaoui showed in When any three solutions are independent, Invent. math. (2022) 230: 1249 \u2013 1265<\/a> that for an ordinary differential field $(k,\\delta)$ of characteristic zero and an irreducible polynomial $P(x_0, x_1, \\ldots, x_n) \\in k[x_0, \\ldots, x_n]$ with $n \\geq 1$, if one can show that for any three distinct solutions $a_1$, $a_2$, and $a_3$ of the differential equation $P(x,\\delta x, \\ldots, \\delta^n x) = 0$ that $a_1$, $a_2$, and $a_3$ are independent in the sense of forking independence or more concretely in the sense that the transcendence degree over $k$ of the field $k(a_1, \\delta a_1, \\ldots, \\delta^{n-1} a_1,\u00a0 a_2, \\delta a_2, \\ldots, \\delta^{n-1} a_2, a_3, \\delta a_3, \\ldots, \\delta^{n-1} a_3)$ is $3n$, then all sets of solutions of that equation are independent.\u00a0 Moreover, they show that if the coefficients of $P$ are constant, then one may replace $3$ by $2$.\u00a0 \u00a0In the follow up work\u00a0 James Freitag, R\u00e9mi Jaoui, Rahim Moosa, The degree of nonminimality is at most 2, Journal of Mathematical Logic, Vol. 23, No. 3 (2023) 2250031 (6 pages)<\/a> the same three authors show that, in fact, relative to $\\operatorname{DCF}_{0,m}$, the theory of differentially closed fields of characteristic zero with $m$ commuting derivations or $\\operatorname{CCM}$, the theory of compact complex manifolds, if $p$ is a stationary type (over any set of parameters) with $U(p) < \\omega$, then the nonminimality degree of $p$ is at most $2$.\u00a0 This theorem is used to improve a theorem of Matthew Delvilbiss and James Freitag\u00a0 that a generic differential equation of high enough degree relative to its order defines a strongly minimal set to the assertion that “high enough degree” means “degree at least six”.<\/p>\n

Some of the key ideas derive from work of Rahim Moosa and Anand Pillay in Some model theory of fibrations and algebraic reductions, Selecta Math. New Ser. (2014) 20:1067 \u2013 1082<\/a> where they gave a model theoretic account of some work on hyperk\u00e4hler manifolds, isolating the notion of a type not admitting a proper fibration.<\/p>\n

For at least the first part of this semester, we will be following the development of the theory of the nonminimality degree through these papers and related work.\u00a0 \u00a0Let me set up some of the details here.<\/p>\n

 <\/p>\n

Minimal types are the building blocks of the finite Lascar rank part of a stable theory.\u00a0 Let us fix some stable theory $T$, a sufficiently large saturated model $\\mathbb{U} \\models T$, and $A \\subsetneq \\mathbb{U}$ a small set of parameters.<\/p>\n

Definition:<\/strong>\u00a0 A type $p \\in S(A)$ is\u00a0algebraic <\/em>if the set of realizations $p(\\mathbb{U})$ is finite.<\/p>\n

Definition:<\/strong> A type $p \\in S(A)$ is\u00a0minimal<\/em> if for every small set $A \\subseteq B \\subsetneq \\mathbb{U}$ there is a unique non-algebraic type $q \\in S(B)$ with $p \\subseteq q$.<\/p>\n

Remark:\u00a0 <\/strong>Note that if $p$ is minimal then it is non-algebraic: take $B := A$, then the only type $q \\in S(B)$ which extends $p$ is $p$ itself.\u00a0 Thus, by definition of minimality it must be that $p = q$ is not algebraic.\u00a0 Likewise, if $p$ is minimal, then it is stationary: consider a set\u00a0 $B$ with\u00a0 $A \\subseteq B \\subsetneq \\mathbb{U}$.\u00a0 Let $q \\in S(B)$ be the unique non-algebraic type extending $p$.\u00a0 \u00a0If $r \\in S(B)$ is any other extension of $p$, then $r$ is algebraic, and, hence, a forking extension of $p$ as $p$ is non-algebraic. By stability, there are non-forking extensions of $p$.\u00a0 Thus, $q$ must be a non-forking extension of $p$ and is the only one.<\/p>\n

By definition, if $p \\in S(A)$ is a non-minimal, non-algebraic stationary type then there is some set $B$ with $A \\subseteq B \\subsetneq \\mathbb{U}$ and a forking extension $q \\in S(B)$ of $p$ with $q$ non-algebraic.\u00a0 Let us consider a Morley sequence $(a_i)_{i=0}^\\infty$ in $q$.\u00a0 As a general fact (See Lemma 2.28 in Pillay’s Geometric Stability Theory), the canonical base of $q$, $\\operatorname{Cb}(q)$, is definable from the Morley sequence.\u00a0 Since $q$ forks over $A$, we have $a_0 \\not \\downarrow_A \\operatorname{Cb}(q)$ which because $\\operatorname{Cb}(q) \\subseteq \\operatorname{dcl}( \\{ a_i : i \\in \\mathbb{Z}_+ \\})$ gives $a_0 \\not \\downarrow_A\u00a0 \\{ a_i : i \\in \\mathbb{Z}_+ \\}$ and then by finite character of forking, we may find a minimal $n \\in \\mathbb{Z}_+$ so that $a_0 \\not \\downarrow_A a_1, \\ldots, a_n$.\u00a0 \u00a0 This observation gives us the degree of nonminimality.<\/p>\n

Definition:<\/strong>\u00a0 If $p \\in S(A)$ is a non-minimal, non-algebraic stationary type then we define the\u00a0nonminimality degree<\/em> of $p$, $\\operatorname{nmd}(p)$, to be the least $n$ for which there are $a_1, \\ldots, a_n$ realizing $p$ for which $p$ has a non-algebraic, forking extension to $A a_1, \\ldots, a_n$.\u00a0 \u00a0 By convention, we set $\\operatorname{nmd}(p) = 0$ for $p$ a minimal or algebraic type.<\/p>\n

Let us consider some examples.<\/p>\n

Example:\u00a0<\/strong> Let $p \\in S(A)$ be a minimal type and let $k > 1$ be any natural number greater than $1$.\u00a0 Let $a_1, \\ldots, a_{k+1}$ be a Morley sequence of length $2k-1$ in $p$ and set $q := p^{\\otimes k} = \\operatorname{tp}(a_1, \\ldots, a_k\/A)$. By the Lascar inequalities $U(q) = k > 1$.\u00a0 \u00a0 Note that $\\operatorname{tp}(a_1, \\ldots, a_k \/ a_2, \\ldots, a_{k+1} A)$ is a non-algebraic forking extension of $q$ over a single realization of $q$.\u00a0 \u00a0Thus, $\\operatorname{nmd}(q) = 1$.<\/p>\n

Let us introduce a defintion.<\/p>\n

Definition:<\/strong> Let $p \\in S(A)$ and $q \\in S(A)$ be two complete types over the same parameter set $A$.\u00a0 We say that $p$ and $q$ are interalgebraic<\/em> if we may find realizations $a$ and $b$ of $p$ and $q$, respectively, so that $\\operatorname{acl}(Aa) = \\operatorname{Ab}$.<\/p>\n

Proposition:\u00a0<\/strong> If $p$ and $q$ are interalgebraic stationary types, then $\\operatorname{nmd}(p) = \\operatorname{nmd}(q)$.<\/p>\n

Proof:\u00a0<\/strong> By the Lascar inequalities, $U(p) = U(q)$.\u00a0 Thus, if we are in the trivial case where $\\operatorname{nmd}(p) =0$ (which is equivalent to $U(p) \\leq 1$) then we also have $\\operatorname{nmd}(q) = 0$.\u00a0 So we may assume that $U(p) > 1$ and $U(q) > 1$.<\/p>\n

Let $a_0, a_1, \\ldots, a_k$ be realizations of $p$ witnessing that $\\operatorname{nmd}(p) \\leq k$.\u00a0 That is, $a_o \\not \\downarrow_A a_1, \\ldots, a_k$ and $a_0 \\notin \\operatorname{acl}(A a_1, \\ldots, a_k)$.\u00a0 By interalgebraicity of $p$ and $q$ we may find $b_0, b_1, \\ldots, b_k$ realizing $q$ so that $\\operatorname{acl}(Aa_i) = \\operatorname{acl}(Ab_i)$ for $0 \\leq i \\leq k$.\u00a0 Thus, $\\operatorname{acl}(A b_1, \\ldots, b_k) = \\operatorname{acl}(A a_1, \\ldots, a_k)$ cannot contain $b_0$, for if it did, because $a_0 \\in \\operatorname{acl}(A b_0)$ and likewise $b_0 \\not \\downarrow_A b_1, \\ldots, b_k$.\u00a0 \u00a0Hence, $\\operatorname{nmd})(q) \\leq \\operatorname{nmd}(p)$.\u00a0 Arguing symmetrically with the roles of $p$ and $q$ reversed, we see that $\\operatorname{nmd}(p) = \\operatorname{nmd}(q)$. $\\Box$<\/p>\n

Examples with $\\operatorname{nmd}(p) > 1$ are hard to come by.\u00a0 Here is one general procedure.<\/p>\n

Construction:\u00a0<\/strong> Let $G \\times X \\to X$ be a $2$-transitive group action definable in some totally transcendental theory.\u00a0 For example, the usual action of $\\operatorname{PGL}_{n+1}(K)$ on $\\mathbb{P}^n(K)$ where $K$ is algebraically closed would fit the bill.\u00a0 Let $p$ be the generic type of $X$ in the two sorted-structure $(G,X)$ having the group structure on $G$ and the action $G \\times X \\to X$.\u00a0 \u00a0Then if $a_1$ and $a_2$ are distinct realizations of $p$ we have that $a_1 \\downarrow a_2$ as the pair $(a_1,a_2)$ has the same type as a pair of indepenent realizations of $p$.\u00a0 If we take $X$ so that $U(p) > 1$ (e.g. take $n \\geq 2$ in the standard example), then we see that $\\operatorname{nmd}(p) \\geq 2$.<\/p>\n

The presence of a group action in the example with $\\operatorname{nmd}(p) > 1$ is not an accident:\u00a0 nonminimality degree is studied through analyses, in the usual stability theoretic sense, and the nontrivial steps in these analyses involve binding groups.<\/p>\n

Definition:<\/strong> A proper fibration<\/em> of the stationary type $p = \\operatorname{tp}(a\/A) \\in S(A)$ is given by some $A$-definable function $f$ on $p$\u00a0 so that\u00a0 $b = f(a) \\in \\operatorname{dcl}(Aa) \\smallsetminus \\operatorname{acl}(A)$ with $a \\notin \\operatorname{acl}(Ab)$.\u00a0 Setting $q := f_*(p) = \\operatorname{tp}(b\/A)$ we have that $q$ is a nonalgebraic type and the fibres of $f$, such as $\\operatorname{tp}(a\/Ab)$, are also nonalgebraic.<\/p>\n

Example:\u00a0<\/strong> If $p \\in S(A)$ and $q \\in S(A)$ are non-algebraic stationary types (possibly equal to each other), then $p \\otimes q$ has a proper fibration.\u00a0 Indeed, let $a \\models p$ and $b \\models q$ with $a \\downarrow_A b$ so that $(a,b) \\models p \\otimes q$.\u00a0 Let $f:p \\otimes q \\to q$ be defined by $(x,y) \\mapsto y$.\u00a0 In particular, $f((a,b)) = b$ and $f_*( (p \\otimes q) ) = q$.\u00a0 We have that $(a,b) \\notin \\operatorname{acl}( A b)$ so that the fiber $\\operatorname{tp}( (a,b) \/ A b)$ is non-algebraic.\u00a0 \u00a0It follows from this example that if $p \\in S(A)$ is any non-algebraic type and $k > 1$, then the Morley power $p^{\\otimes k}$ has a proper fibration.<\/p>\n

Proposition:\u00a0<\/strong> If $p$ has a proper fibration, then $\\operatorname{nmd}(p) = 1$.<\/p>\n

Proof:\u00a0 Let $a$ realize $p$ and $b \\in \\operatorname{dcl}(Aa) \\smallsetminus \\operatorname{acl}(A)$ witness that $p$ has a proper fibration. Let $f:p \\to q := \\operatorname{tp}(b\/A)$ be an $A$-definable function with $f(a) = b$.\u00a0 Let $a’$ be any realization of $\\operatorname{tp}(a\/Ab)$ which is not\u00a0 in $\\operatorname{acl}(Aa)$. This is possible as $\\operatorname{tp}(a\/Ab)$ is not algebraic.\u00a0 \u00a0Since $b \\in \\operatorname{dcl}(Aa) \\smallsetminus \\operatorname{acl}(A)$, we have $b \\not \\downarrow_A a$.\u00a0 As $b \\in \\operatorname{dcl}(Aa’)$, we have $a’ \\not \\downarrow_A a$.\u00a0 That is, $\\operatorname{tp}(a’\/Aa)$ is a forking extension of $\\operatorname{tp}(a’\/A) = p$.\u00a0 As $a’ \\notin \\operatorname{acl}(Aa)$, this type is nonalgebraic.\u00a0 That is, we have succeeded in finding a nonalgebraic forking extension of $p$ over a single realization of $p$; put another way: $\\operatorname{nmd}(p) = 1$. $\\Box$<\/p>\n

Read contrapositively, a stationary type $p$ with $\\operatorname{nmd}(p) > 1$ cannot have a proper fibration.<\/p>\n

The main result of the Moosa-Pillay paper mentioned above descibes when a stationary type of finite Lascar rank has no proper fibration.<\/p>\n

Theorem:<\/strong>\u00a0 If the $p$ is a stationary type with $0 < U(p) < \\omega$ and $p$ does not have a proper fibration, then either there is a modular minimal type $q$ for which $p$ is interalgebraic with $q^{\\otimes U(p)}$ or there is a non-locally modular type $r$ for which $p$ is almost $r$-internal.<\/p>\n

We defer the proof of this theorem to a later lecture.<\/p>\n

Combining the Moosa-Pillay theorem with the proposition relating proper fibrations to nonminimality degree, we obtain the following proposition.<\/p>\n

Proposition:<\/strong> If $p \\in S(A)$ is stationary with $U(p) < \\omega$ and $\\operatorname{nmd}(p) > 1$, then there is nonlocally modular minimal type $r$ so that $p$ is almost $r$-internal.<\/p>\n

Proof:<\/strong> By the proposition on fibrations and nonminimality degree, $p$ has no proper fibration.\u00a0 By the Moosa-Pillay theorem, either there is a minimal type $r$ so that either (1) $r$ is modular and $p$ is interalgebraic with $r^{\\otimes U(p)}$ or (2) $r$ is non-locally modular and $p$ is almost $r$-internal.\u00a0 \u00a0We need to rule out the first case.\u00a0 \u00a0Since $\\operatorname{nmd}(p) > 1$, we must have $U(p) > 1$. By our observations on Morley products and fibrations, $r^{\\otimes U(p)}$ has a proper fibration.\u00a0 Then by the proposition on proper fibrations and nonminimality degree, $\\operatorname{nmd}(r^{\\otimes U(p)}) = 1$.\u00a0 As $p$ and $r^{\\otimes U(p)}$ are interalgebraic, it would follow that $\\operatorname{nmd}(p) = \\operatorname{nmd}(r^{\\otimes U(p)}) = 1$, contrary to our hypothesis on $p$.\u00a0 Therefore, we must be in case (2).\u00a0 $\\Box$<\/p>\n

Remark:\u00a0<\/strong> In later lectures, we will see how by replacing $p$ with a type $q$ which interalgebraic with $p$, we may upgrade from almost internality<\/em> to a non-locally modular minimal type to outright internality<\/em>.<\/p>\n

Granting this, to study the case where $U(p) < \\omega$ and $\\operatorname{nmd}(p) > 1$, we may assume that $p$ is internal to some non-locally modular minimal type $r$.<\/p>\n

Proposition:<\/strong>\u00a0 If $p \\in S(A)$ is stationary,\u00a0 $r \\in S(A)$ with<\/p>\n