Fact 2.6 [Reference d’Elbée, Müller, Ramsey and Siniora10, Proposition 4.22, Corollary 4.40].

The relation satisfies symmetry, invariance, monotonicity, base monotonicity, stationarity, transitivity, full existence (for all $A,B,C$ there exists $A'\equiv _C A $ such that ). In other words, is a stationary independence relation in the sense of [Reference Tent and Ziegler25].

Proof. We prove by induction on terms of $\mathscr {L}_{K,V,c}$ that if $\overline {\alpha }$ is a tuple of scalars from $K'$ and $\overline {v}$ is a tuple of vectors from V, then $t(\overline {\alpha }, \overline {v}) \in K'$ or $t(\overline {\alpha },\overline {v}) = \sum _{i < k} \beta _{i} w_{i}$ for $\beta _{i} \in K'$ and $w_{i} \in V$ for some k. Since the expression $\beta w$ in $V'$ can be identified with $\beta \otimes w$ in $K' \otimes _{K} V$ , this suffices. The desired conclusion is clear for terms t consisting of variables and constants, so the base case is true. It is also clear that the conclusion is preserved by application of the field operations (or, more generally, the operations in $L^{\dagger }$ ) and the vector space operations. Bilinearity of the bracket entails that

$$ \begin{align*}\left[\sum_{i < k}\beta_{i} w_{i}, \sum_{j < k'}\beta^{\prime}_{j}w^{\prime}_{j}\right] = \sum_{\substack{i < k\\j < k'}} (\beta_{i} \beta^{\prime}_{j})[w_{i},w^{\prime}_{j}], \end{align*} $$

where $\beta _{i} \beta ^{\prime }_{j} \in K'$ and $[w_{i},w^{\prime }_{j}] \in V$ , so the conclusion is also preserved by the bracket. Finally, we check the coordinate functions. We will proceed as in [Reference Kestner and Ramsey17, Lemma 1.5]. Suppose we are given $v_{1}, \ldots , v_{n+1}$ of the form

$$ \begin{align*}v_{i} = \sum_{j = 1}^{m} \alpha_{i,j} w_{j} \end{align*} $$

for vectors $w_{1}, \ldots , w_{m} \in V$ and $\alpha _{i,j} \in K'$ . Note that, as some $\alpha _{i,j}$ are allowed to be zero, we may assume that the vectors $w_{j}$ that appear in the above expression are the same for all $v_{i}$ . We assume $v_{1}, \ldots , v_{n}$ are linearly independent and $v_{n+1}$ is in their span, so $v_{n+1} = \sum _{i = 1}^{n} \lambda _{i} v_{i}$ . Thus, replacing the $w_{j}$ ’s with a subset, we may assume that the $w_{j}$ are linearly independent. Further, after possibly extending the set of $v_i$ ’s by linearly independent $w_j$ ’s, we may consider $m = n$ . As $\lambda _{i} = \pi _{n,i}(v_{1}, \ldots , v_{n+1})$ , we must show that each $\lambda _{i} \in K'$ . Writing matrices with respect to the basis $w_{1}, \ldots , w_{n}$ , we have

$$ \begin{align*}\left[ \begin{matrix} \alpha_{1,1} & \ldots & \alpha_{n,1} \\ \vdots & \ddots & \\ \alpha_{1,n} & \dots & \alpha_{n,n} \end{matrix} \right] \left[ \begin{matrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{matrix} \right] = \left[ \begin{matrix} \alpha_{n+1,1} \\ \vdots \\ \alpha_{n+1,n} \end{matrix} \right]. \end{align*} $$

Writing $B = (\alpha _{i,j}) \in \mathrm {GL}_{n}(K')$ , we see that

$$ \begin{align*}B^{-1} \left[ \begin{matrix} \alpha_{n+1,1} \\ \vdots \\ \alpha_{n+1,n} \end{matrix} \right] = \left[ \begin{matrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{matrix} \right], \end{align*} $$

which shows that indeed each $\lambda _{i} \in K'$ .

Proof. If $M = (K(M), V(M), [\cdot ,\cdot ]^{M}) {\vDash } T_{0}$ , then, by extension of scalars, we can assume $K(M)$ is an algebraically closed field. Hence, $T = T^{+}$ for the special case that $\mathscr {L}^{\dagger } = \mathscr {L}_{\mathrm {rings}}$ and $T^{\dagger } = ACF$ . Therefore, it suffices to argue that if $M = (K(M), V(M), [\cdot ,\cdot ]^{M}) {\vDash } T_{0}^{+}$ , then M embeds into a model of $T^{+}$ .

Let $(\overline {\alpha }_{i})_{i}$ list all possible finite sequences of scalars from $K(M)$ which are the structure constants of a finite dimensional c-nilpotent LLA $L_{i}$ over $K(M)$ . Let $V' = V \oplus \bigoplus _{i} L_{i}$ and let $[\cdot ,\cdot ]' : V' \times V' \to V'$ be the associated direct sum Lie bracket. Then $M' = (K(M), V', [\cdot ,\cdot ]')$ satisfies the axiom schema (1) and (2) in the definition of $T^{+}$ .

To satisfy the axiom scheme (3), we will build the model as the union of a chain. Let $M_{0} = M'$ be the model as constructed above. Now, given $M_{i}$ , we will construct an extension $M_{i+1}$ such that if $\overline {v}$ is a basis of a finite dimensional subalgebra of $(V(M_{i}),[\cdot ,\cdot ]^{M_{i}})$ and $\overline {\alpha }$ is a finite sequence of structure constants for some c-nilpotent LLA extending the subalgebra spanned by $\overline {v}$ , then there is some $\overline {w}$ in $V(M_{i+1})$ such that $\overline {v}$ and $\overline {w}$ together form a basis for a subalgebra of $(V(M_{i+1}),[\cdot ,\cdot ]^{M_{i+1}})$ with structure constants $\overline {\alpha }$ . Let $(L_{j},\overline {\alpha }_{j})_{j < \lambda }$ list all pairs where $L_{j}$ is a finite dimensional subalgebra of $(V(M_{i}),[\cdot ,\cdot ]^{M_{i}})$ and $\overline {\alpha }_j$ is a finite sequence of structure constants for some c-nilpotent LLA extending $L_{j}$ . For each j, let $L^{\prime }_{j}$ be an LLA extending $L_{j}$ with structure constants $\overline {\alpha }_{j}$ .

We define $N_{0}$ to be the LLA $(V(M_{i}), [\cdot ,\cdot ]^{M_{i}})$ . Then given $N_{j}$ for some $j < \lambda $ , we define $N_{j+1}$ to be the free amalgam of $N_{j}$ and $L^{\prime }_{j}$ over $L_{j}$ , as LLAs over $K(M_{i})$ . Note that this exists by [Reference d’Elbée, Müller, Ramsey and Siniora10, Theorem 4.35] (where it is denoted $N_{j} \otimes _{L_{j}} L^{\prime }_{j}$ , viewing them as LLAs over $K(M_{i})$ ). We view $N_{j+1}$ as an extension of $N_{j}$ . As for limit ordinals $\delta \leq \lambda $ , we define $N_{\delta } = \bigcup _{j < \delta } N_{j}$ . Then we set $M_{i+1} = (K(M_{i}), N_{\lambda }, [\cdot ,\cdot ]^{N_{\lambda }})$ .

To conclude we set $M" = \bigcup _{i < \omega } M_{i}$ . Axiom scheme (1) is clearly satisfied. Moreover, since $M'$ satisfies the second axiom scheme and $M"$ has the same set of scalars, $M"$ satisfies it as well. Finally, if L is a finite dimensional subalgebra of $M"$ and $\overline {\alpha }$ is a sequence of structure constants for a finite dimensional extension of L, then L is contained in $M_{i}$ for some i and hence a extension of L with structure constants $\overline {\alpha }$ can be found as a subalgebra of $M_{i+1}$ , hence of $M"$ . This shows the axiom scheme (3) is satisfied as well, completing the proof.

Proof. We first argue for T. We use the well-known criterion for QE that if M and N are two $\aleph _{0}$ -saturated models of T whose fields have the same characteristic, then the set of partial isomorphisms from a finitely generated substructure of M to a finitely generated substructure of N has the back-and-forth property. Non-emptiness is easy, since if $K_{0}$ is the prime subfield then the unique map $f:(K_{0},\left \{ {0} \right \})\to (K_{0},\left \{ {0} \right \})$ is a partial isomorphism.

Now suppose $M_{0} \subseteq M$ and $N_{0} \subseteq N$ are finitely generated substructures and $f : M_{0} \to N_{0}$ is an isomorphism. Suppose $a \in M \setminus M_{0}$ . Let $M_{1} = \langle a M_{0} \rangle $ . We have to consider three cases:

Case 1. $a \in K(M)$ .

Because $K(N)$ is algebraically closed, we can find some b such that the isomorphism from $K(M_{0})$ to $K(N_{0})$ given by f extends to an isomorphism from $K(M_{1})$ , i.e., the subfield of $K(M)$ generated by $K(M_{0})a$ , to the subfield of $K(N)$ generated by $K(N_{0})b$ and mapping $a \mapsto b$ . Let $N_{1} = \langle N_{0},b \rangle $ . Then, by Lemma 2.7, $(V(M_{1}), [\cdot ,\cdot ])$ and $(V(N_{1}), [\cdot ,\cdot ])$ are obtained by extension of scalars from $K(M_{0})$ to $K(M_{1})$ and from $K(N_{0})$ to $K(N_{1}),$ respectively. The isomorphism $K(M_{1})$ to $K(N_{1})$ extending f, then, induces an isomorphism from $V(M_{1})$ to $V(N_{1})$ respecting the Lie algebra structure, as desired.

Case 2. $a \in V(M)$ and a is in the $K(M)$ -span of $V(M_{0})$ .

Choose a basis $\overline {v}$ for $V(M_{0})$ . If a is in the $K(M)$ -span of $\overline {v} = (v_{0}, \ldots , v_{n-1})$ then, since $a \not \in V(M_{0})$ , it must be the case that

$$ \begin{align*}a = \sum_{i < n} \alpha_{i} v_{i} \end{align*} $$

with not all $\alpha _{i} \in K(M_{0})$ . Applying Case 1 at most n times, we may extend f to an isomorphism $f': \langle M_{0}, \alpha _{<n} \rangle \to \langle N_{0}, \beta _{<n} \rangle $ where $f'(\alpha _{i}) = \beta _{i}$ for all $i < n$ . Note that then

$$ \begin{align*}f'(a) = \sum_{i < n} \beta_{i}f(v_{i}). \end{align*} $$

Because we have the coordinate functions in the language, we have $M_{1} = \langle M_{0}, \alpha _{<n}\rangle $ and, setting $b = f'(a)$ , we also have $N_{1} = \langle N_{0},b \rangle = \langle N_{0}, \beta _{<i} \rangle $ and $f' :M_{1} \to N_{1}$ is the desired extension.

Case 3. $a \in V(M)$ and a is not in the $K(M)$ -span of $V(M_{0})$ .

We may assume that $[a,u]$ is in the $K(M)$ -span of $V(M_{0})$ for all $u \in V(M_{0})$ (i.e, that the Lie algebra over $K(M)$ spanned by $V(M_{0})$ is an ideal of the Lie algebra over $K(M)$ spanned $V(M_{1})$ ) since the general case follows from this one by reverse induction on the maximum $1\leq i\leq c+1$ such that $a\in P_i(M)\setminus P_{i+1}(M)$ . Let $\{v_1,\dots , v_n\}$ be a $K(M_0)$ -basis of $M_0$ and let $M_{0}^{\prime }$ be the substructure of M generated by $M_{0}$ and $\{[a,v_{i}] : i < n\}$ . Notice that $\overline {v}$ is still a basis of $V(M_{0}^{\prime })$ though the field may grow. By at most n applications of Case 2, there is a structure $N_{0} \subseteq N_{0}^{\prime } \subseteq N$ and an isomorphism $f': M_{0}^{\prime } \to N_{0}^{\prime }$ extending f.

Let $\overline {\alpha }$ be the sequence of structure constants for the subalgebra of M spanned by $\overline {v}$ and a. By construction, $\overline {\alpha }$ is contained in $K(M_{0}^{\prime })$ and hence $f'(\overline {\alpha })$ is contained in $K(N_{0}^{\prime })$ . Since $f'(\overline {\alpha })$ is a sequence of structure constants for an LLA extending $f'(\overline {v})$ , the axioms of T entail that there is some $b \in V(N)$ such that $f'(\overline {v})$ and b span an LLA with structure constants $f'(\overline {\alpha })$ . Then the map extending $f'$ and mapping $a \mapsto b$ defines an isomorphism from $M_{1} = \langle M_{0}^{\prime }a \rangle $ to $\langle N_{0}^{\prime }b \rangle $ , which gives the desired extension.

The proof in the case that we are considering $T^{+}$ is essentially identical, with minor modifications. Instead of considering all partial isomorphisms, we only consider those partial isomorphisms in which the induced $L^\dagger $ -isomorphism of fields $K(M_0)\to K(N_0)$ is $T^\dagger $ -elementary. Then, since $K(M)$ and $K(N)$ are $\aleph _{0}$ -saturated as models of $T^{\dagger }$ , this map may be extended to incorporate a new field element, which yields Case 1. Case 2, and 3 are identical.

Proof. Immediately follows from $(\star )$ .

Proof. Immediate.

Proof. Let $f : A_{0} \to A_{1}$ be a partial elementary isomorphism fixing B pointwise. So we may write $f = (f_{K},f_{V})$ where $f_{K} : K(A_{0}) \to K(A_{1})$ fixes $K(B)$ and $f_{V}: V(A_0) \to V(A_1)$ fixes $V(B)$ . Since $K(A_{0}) \equiv _{K'} K(A_{1})$ , we know $f_{K}$ extends to a partial elementary map $g: K(A_{0})K' \to K(A_{1})K'$ which fixes $K'$ pointwise. Note that by Lemma 2.7, $\langle A_{i},K'\rangle = ((K(A_{i})K'),K(A_{i})K' \otimes _{K(A_{i})} V(A_{i}))$ for $i = 0,1$ . We can define an isomorphism $h : \langle A_{0},K' \rangle \to \langle A_{1},K' \rangle $ with $h = (h_{K},h_{V})$ where $h_{K} = g$ and $h_{V} : (K(A_{0})K') \otimes _{K(A_{0})} V(A_{0}) \to (K(A_{1})K') \otimes _{K(A_{1})} V(A_{1})$ by setting

$$ \begin{align*}h(\alpha \otimes v) = g(\alpha) \otimes f_{V}(v), \end{align*} $$

for all $\alpha \in K(A_{0})K'$ and $v \in V(A)$ and extending linearly. Since $h_{K} = g$ is partial elementary in the field sort, this isomorphism is partial elementary. Since h fixes B (as it extends f) and fixes $K'$ (by the choice of g), the map h witnesses $A_{0} \equiv _{BK'} A_{1}$ .

Observation 3.6. Suppose $M = (K,V), M' = (K',V') {\vDash } T^{+}$ are countable models and $K \cong K'$ . Then $M \cong M'$ .

Proof. An easy back-and-forth.

Proof. We may replace M with a very saturated elementary extension $\mathbb {M}$ . Suppose A is a substructure of $\mathbb {M}$ with $\mathrm {acl}_{L^{\dagger }}(K(A)) = K(A)$ and let B be a substructure with $A \subseteq B \subseteq \mathbb {M}$ . Let $(K_{i})_{i < \omega }$ be a $K(A)$ -indiscernible seqeunce in $K(\mathbb {M})$ with $K_{0} = K(B)$ and for all $i < \omega $ , where is computed in $T^{\dagger }$ . Note that, by quantifier-elimination, we have that $(K_{i})_{i < \omega }$ is A-indiscernible in $\mathbb {M}$ . Let $B_{0} = B$ and choose, for each $i> 0$ , some $B_{i}$ with $K(B_{0}) B_{0} \equiv _{A} K_{i} B_{i}$ . In particular, we have $K(B_{i}) = K_{i}$ for all i.

Let $\tilde {K} = K(\langle B_{i} : i < \omega \rangle )$ . Let $\tilde {A}$ and $\tilde {B}_{i}$ denote the respective extensions of scalars of A and $B_{i}$ to $\tilde {K}$ —that is, $\tilde {A} = \langle \tilde {K}, A\rangle $ and $\tilde {B}_{i} = \langle \tilde {K}, B_{i} \rangle $ for all $i < \omega $ . Then, inductively, we define a sequence $(\tilde {B}^{\prime }_{i})_{i < \omega }$ by setting $\tilde {B}^{\prime }_{0} = \tilde {B}_{0}$ and, given $\tilde {B}_{0}, \ldots , \tilde {B}_{i}$ , we choose $\tilde {B}_{i+1}' \equiv _{\tilde {A}} \tilde {B}_{i+1}$ with as LLAs over $\tilde {K}$ . Choose, for each $i> 0$ , some $\sigma _{i} \in \mathrm {Aut}(\mathbb {M}/\tilde {A})$ such that $\sigma (\tilde {B}_{i}) = \tilde {B}^{\prime }_{i}$ and define $B^{\prime }_{i} = \sigma (B_{i})$ . Since each $\sigma _{i}$ fixes $\tilde {A}$ , it fixes $\tilde {K}$ and thus fixes $K(B_{i})$ . It follows that $K(B^{\prime }_{i}) = K(B_{i})$ . We also have, by free amalgamation, that $V(B_{i})$ is linearly independent from $V(B^{\prime }_{\leq i})$ over $V(A)$ . By quantifier elimination, we have $B^{\prime }_{i} \equiv _{A} B$ and, by construction, $B^{\prime }_{i} \cap B^{\prime }_{j} = A$ for all $i \neq j$ . This shows that A is algebraically closed.

Proof. Let $\tilde {M}$ denote the ultraproduct $\prod M_{p}/\mathcal {D}$ . We have that $\mathbf {L}_{c,p}$ and $M_{p}$ are two ways of encoding the same Lie algebra (though $M_{p}$ has as its underlying set the disjoint union of the Lie algebra $\mathbf {L}_{c,p}$ and the field $\mathbb {F}_{p}$ ). For each $k\in \mathbb {N}$ , and each k-generated LLA A over $\mathbb {F}_{p}$ , there is a formula $\varphi _{A}(x_{1}, \ldots , x_{k})$ such that, for any arbitrary $N_{p}$ , $N_{p} {\vDash } \varphi (a_{1}, \ldots , a_{k})$ if and only if $\langle a_{1}, \ldots , a_{k} \rangle \cong A$ .

Because $\mathbf {L}_{c,p}$ is the Fraïssé limit of the class of all finite c-nilpotent LLAs over $\mathbb {F}_{p}$ , the theory $\mathrm {Th}(\mathbf {L}_{c,p})$ is axiomatized by saying the following.

1. (1) For each finite c-nilpotent LLA A over $\mathbb {F}_{p}$ generated by a basis of k elements, there is an axiom $(\exists x_{1}, \ldots , x_{k})\varphi _{A}(x_{1}, \ldots , x_{k})$ . This expresses that $N_{p}$ has as its age all finite c-nilpotent LLAs over $\mathbb {F}_{p}$ .

2. (2) For each containment $A \subsetneq B$ of finite LLAs over $\mathbb {F}_{p}$ , with A k-dimensional and B l-dimensional, there is an axiom

   $$ \begin{align*}(\forall x_{1}, \ldots, x_{k})(\exists y_{k+1}, \ldots, y_{l})[\varphi_{A}(\overline{x}) \to \varphi_{B}(\overline{x}, \overline{y})]. \end{align*} $$

These axioms express that $\mathbf {L}_{c,p}$ satisfies the embedding property and hence is homogeneous. Now we just need to check that $\tilde {M}$ satisfies the axiom scheme (1)–(3) at the beginning of Section 3. Axiom scheme (1) is clear since $T^{\dagger }$ is the theory of pseudo-finite fields and each $\mathbf {L}_{p,c}$ is clearly infinite dimensional. Axiom (2) and (3) follow by a standard application of Łos’s theorem. As (3) is entirely similar, we will only spell out the details for (2).

Let $\tilde {K} = K(\tilde {M}) = \prod \mathbb {F}_{p}/\mathcal {D}$ . Suppose $\overline {\alpha } = (\alpha _{1},\ldots , \alpha _{n^{3}}) \in \tilde {K}$ and we have

$$ \begin{align*}\tilde{M} {\vDash} \varphi_{\mathrm{str},n}(\overline{\alpha}). \end{align*} $$

Write $\overline {\alpha } = (\alpha _{1,p}, \ldots , \alpha _{n^{3},p})_{p}/\mathcal {D}$ . For each prime p, if $M_p{\vDash } \varphi _{\mathrm {str},n}(\alpha _{1,p}, \ldots , \alpha _{n^{3},p})$ , we will let $A_{p} \subseteq \mathbf {L}_{p,c}$ be a Lie algebra over $\mathbf {F}_{p}$ with structure constants $(\alpha _{1,p}, \ldots , \alpha _{n^{3},p})$ , which exists by the axiom scheme (1) above. Then we can choose $\overline {v}$ from $A_{p}$ such that $(\alpha _{1,p}, \ldots , \alpha _{n^{3},p})$ are structure constants for $A_{p}$ with respect to the basis $v_{1,p}, \ldots , v_{n,p}$ . If $(\alpha _{1,p}, \ldots , \alpha _{n^{3},p})$ are not structure constants, then we set $A_{p} = 0$ (and likewise each $v_{i,p} = 0$ ). By Łos’s theorem, $A_{p} \neq 0$ for all but a $\mathcal {D}$ -small set of primes. Setting $(v_{1},\ldots , v_{n}) = (v_{1,p},\ldots , v_{n,p})_{p}/\mathcal {D}$ , we have, again by Łos’s theorem, that

$$ \begin{align*}\tilde{M} {\vDash} \varphi_{\mathrm{Lie},n}(v_{1},\ldots, v_{n}) \wedge \mathrm{Str}_{n}(\overline{v}) = \overline{\alpha}. \end{align*} $$

As $\overline {\alpha }$ was an arbitrary $n^{3}$ -tuple from $\tilde {K}$ , we can conclude that $\tilde {M}$ satisfies axiom scheme (2), completing the proof.

Fact 4.1 [Reference Chernikov and Hempel7].

Let M be an $\mathscr {L}'$ -structure such that its reduct to a language $\mathscr {L} \subseteq \mathscr {L}'$ is NIP. Let $d,k$ be natural numbers and $\varphi (x_{1}, \ldots , x_{d})$ be an $\mathscr {L}$ -formula. Let further $y_{0}, \ldots , y_{k}$ be arbitrary $(k+1)$ -tuples of variables. For each $1 \leq t \leq d$ , let $0 \leq i_{t,1}, \ldots , i_{t,k} \leq k$ be arbitrary and let $f_{t} : M_{y_{i_{t,1}}} \times \ldots \times M_{y_{i_{t,k}}} \to M_{x_{t}}$ be an arbitrary k-ary function. Then the formula

$$ \begin{align*}\psi(y_{0}; y_{1}, \ldots, y_{k}) = \varphi(f_{1}(y_{i_{1,1}}, \ldots, y_{i_{1,k}}), \ldots, f_{d}(y_{i_{d,1}}, \ldots, y_{i_{d,k}})) \end{align*} $$

is k-dependent.

Proof. (1) It is clear that the statement is true for variables and constants. It is also clear that the class of terms of this form is closed under scalar multiplication by a field valued term $s(\overline {x},\overline {y})$ , vector addition and subtraction, and by application of the Lie bracket. Therefore, the conclusion follows by induction on terms.

(2) We will argue by induction on terms using (1). The conclusion is obvious for constants and variables. The class of terms satisfying the description in (2) is also clearly closed under the field operations. The only non-trivial case, then, is to check that a coordinate function applied to vector space sort valued terms can again be written in this form. Fix natural numbers $i_{*} \leq n$ . So assume $t_{1}(\overline {x},\overline {y}), \ldots , t_{n+1}(\overline {x},\overline {y})$ are vector space sort valued terms and

$$ \begin{align*}t(\overline{x},\overline{y}) = \pi_{n,i_{*}}(t_{1}(\overline{x},\overline{y}), \ldots, t_{n+1}(\overline{x},\overline{y})), \end{align*} $$

is a term of minimal complexity for which the conclusion has not yet been established. By (1), for each $1 \leq i \leq n+1$ , we have

$$ \begin{align*}t_{i}(\overline{x},\overline{y}) = \sum_{j = 1}^{k_{i}} t_{i,j}(\overline{x},\overline{y})m_{i,j}'(\overline{y}), \end{align*} $$

where each $t_{i,j}$ is a term valued in the field sort and $m^{\prime }_{i,j}(\overline {y})$ is a Lie monomial. By induction, then, each term $t_{i,j}(\overline {x},\overline {y}) = s_{i,j}(\overline {x}, \overline {m}_{i,j}(\overline {y}))$ , where $s_{i,j}(\overline {x},\overline {z}_{i,j})$ is a term of $\mathscr {L}_{K,V,c}^{-}$ and $\overline {m}_{i,j}(\overline {y})$ is a tuple of Lie monomials. Therefore, we have

$$ \begin{align*}t(\overline{x},\overline{y}) = \pi_{n,i_{*}}\left( \sum_{j = 1}^{k_{1}} s_{1,j}(\overline{x},\overline{m}_{1,j}(\overline{y}))m^{\prime}_{1,j}(\overline{y}), \ldots, \sum_{j = 1}^{k_{n+1}} s_{n+1,j} (\overline{x},\overline{m}_{n+1,j}(\overline{y}))m^{\prime}_{n+1,j}(\overline{y})\!\right). \end{align*} $$

Consider the $\mathscr {L}_{K,V,c}^{-}$ term $s(\overline {x},\overline {z},\overline {w})$ defined by

$$ \begin{align*}s(\overline{x},\overline{z},\overline{w}) = \pi_{n,i_{*}}\left( \sum_{j = 1}^{k_{1}} s_{1,j}(\overline{x},\overline{z}_{1,j})w_{i,j}, \ldots, \sum_{j = 1}^{k_{n+1}} s_{n+1,j} (\overline{x},\overline{z}_{n+1,j})w_{i,j}\right), \end{align*} $$

where $\overline {z}$ is a tuple concatenating the tuples $\overline {z}_{i,j}$ and $\overline {w}$ is a tuple enumerating the $w_{i,j}$ . Then if $\overline {m}(\overline {y})$ concatenates the tuples $\overline {m}_{i,j}(\overline {y})$ and $\overline {m}'(\overline {y})$ enumerates the $m^{\prime }_{i,j}(\overline {y})$ , then we obtain

$$ \begin{align*}t(\overline{x},\overline{y}) = s(\overline{x},\overline{m}(\overline{y}),\overline{m}'(\overline{y})), \end{align*} $$

which has the desired form. This completes the induction.

Proof. The reduct of T to the language $\mathscr {L}^{-}_{K,V,c}$ is the theory of an infinite dimensional vector space over an algebraically closed field, with a distinguished flag of length c such that each quotient $P_{i}(M)/P_{i+1}(M)$ is infinite-dimensional. It is complete after specifying a characteristic of the field and a model M of size $\aleph _{1}$ of a completion is determined by specifying the transcendence degree of the field over the prime subfield, and the dimension of each quotient $P_{i}(M)/P_{i+1}(M)$ . There are countably many choices for the transcendence degree, since this can be finite, $\aleph _{0}$ , or $\aleph _{1}$ , and there are two choices, $\aleph _{0}$ or $\aleph _{1}$ , for each quotient. It follows that there are only $\aleph _{0}$ many models of size $\aleph _{1}$ . This implies that every completion of the reduct of T to the language $\mathscr {L}^{-}_{K,V,c}$ is $\omega $ -stable.

Proof. The proof that T is $(c-1)$ -independent is identical to the argument of [Reference d’Elbée, Müller, Ramsey and Siniora10, Theorem 5.20], replacing the free c-nilpotent Lie algebra over $\mathbb {F}_{p}$ with the free c-nilpotent Lie algebra over any algebraically closed field. Moreover, it follows by Lemma 4.2 that every formula $\varphi (\overline {x},\overline {y})$ of $\mathscr {L}_{K,V,c}$ , where $\overline {x}$ is a tuple of field sort variables and $\overline {y}$ is a tuple of vector space sort variables, can be written in the form

$$ \begin{align*}\psi(\overline{x},\overline{m}(\overline{y})) \end{align*} $$

for an $\mathscr {L}_{K,V,c}^{-}$ -formula $\psi (\overline {x},\overline {z})$ , where $\overline {m}(\overline {y})$ is a tuple of Lie monomials in the variables $\overline {y}$ . By c-nilpotence, every Lie monomial is at most c-ary. By Lemma 4.3, the formula $\psi (\overline {x},\overline {z})$ is stable. Therefore, by Fact 4.1, we conclude that T is c-dependent.

Fact 4.8. Suppose $(a_i)_{i<\omega }$ is an indiscernible sequence in $\mathbb {M}$ . Then there is a model M such that $(a_i)_{i<\omega }$ is -independent over M.

Proof. Apply [Reference d’Elbée9, Theorem 3.3.7].

Fact 4.11 (Strengthened independence theorem [Reference Kruckman and Ramsey18, Theorem 2.13]).

Assume that T is NSOP $_1$ and $M\prec \mathbb {M}$ , , , and $c_1\equiv _M c_2$ . Then there exists c such that $c\equiv _{Ma}c_1$ , $c\equiv _{Mb} c_2,$ and $(a,b,c)$ is an independent triple over M (by which we mean , and ).

Fact 4.12. If T is NSOP $_1$ and every small subset of $\mathbb {M}$ is an extension base for forking, then for all $a,b,C$ there exists $a'\equiv _C a$ such that .

Fact 4.13 [Reference Kaplan and Ramsey15, Proposition 9.28] [Reference Chatzidakis4].

Let T be an NSOP $_1$ theory of fields, then implies .

Proof. Let $I = (B_{i})_{i < \omega }$ be an indiscernible sequence with $B_{0} = B$ which is both tree Morley over M and over $N_{1}$ , which exists by [Reference Kaplan and Ramsey16, Proposition 6.4]. By the chain condition, there is $I' \equiv _{MB_{0}} I$ such that and that $I'$ is $MA$ -indiscernible. Let $N_{1}^{\prime }$ be a model with $N^{\prime }_{1}I' \equiv _{MB_{0}} N_{1}I$ . Note that, by invariance, we have and $I'$ is tree Morley over M. Thus, we have and, hence, by the independence theorem, there is some $N_{*} {\vDash } \mathrm {tp}(N_{0}/MA) \cup \mathrm {tp}(N^{\prime }_{1}/MI')$ with . Then $I'$ is both $N_{*}$ -indiscernible and $MA$ -indiscernible. By Ramsey and compactness, extract an $N_{*}A$ -indiscernible sequence $I"$ from $I'$ . Then $I"$ is also a tree Morley sequence over $N_{*}$ and $I" \equiv _{MA} I'$ . Choose some model N such that $NI' \equiv _{MA} N_{*}I"$ . Thus, $I'$ is $NA$ -indiscernible and is a tree Morley sequence over N starting with B. Thus, by witnessing, we have .

Proof. We may view $A' \otimes _{C'} B'$ as the sub-LLA over $K'$ of M generated by $A'$ and $B'$ . In the two-sorted language, this amounts to saying that $A' \otimes _{C'} B'$ can be identified with the vector space sort of $\langle A',B' \rangle = \langle A,B,K' \rangle $ in $M'$ . Let $\tilde {A} = \langle A,K\rangle $ , $\tilde {B} = \langle B,K \rangle $ , and $\tilde {C} = \langle C,K\rangle $ be the extensions of scalars of A, B, and C to K. We will argue that $\langle A,B \rangle $ can be identified with $\tilde {A} \otimes _{\tilde {C}} \tilde {B}$ .

Let L be any LLA over K and consider LLA homomorphisms $f: \tilde {A} \to L$ and $g: \tilde {B} \to L$ over K which agree on $\tilde {C}$ . Let $L'$ be the extension of scalars of L to $L'$ , an LLA over $K'$ . Then f and g induce $K'$ -linear maps $f' : A' \to L'$ and $g' : B' \to L'$ which agree on $C'$ . Hence, there is a map $h' : \langle A',B' \rangle \to L'$ extending $f'$ and $g'$ . In particular, $h = h'|_{\langle A,B \rangle }$ extends f and g. Since the image of h is, therefore, generated as an LLA over K by $f(\tilde {A})$ and $g(\tilde {B})$ , it follows that the image of h lands in L. Applying this to the $L = \tilde {A} \otimes _{\tilde {C}} \tilde {B}$ with $f= \mathrm {id}_{\tilde {A}}$ and $g = \mathrm {id}_{\tilde {B}}$ , we see in particular that $K(\langle A,B \rangle ) = K$ . More generally, we have shown $\langle A,B \rangle $ satisfies the desired universal property so $\langle A,B \rangle \cong \tilde {A} \otimes _{\tilde {C}} \tilde {B}$ .

Proof. Invariance and symmetry are clear, by the invariance and symmetry of in NSOP $_{1}$ theories. To prove full existence, suppose we are given algebraically closed A, B, and C with $C \subseteq A \cap B$ . We will assume $K(C)$ is a model of $\mathrm {Th}(K(\mathbb {M}))$ . We want to find some $D \equiv _{C} A$ such that . First, we will use full existence in $\mathrm {Th}(K(\mathbb {M}))$ to find some $K' \equiv _{K(C)} K(A)$ with . Let $\sigma \in \mathrm {Aut}(K(\mathbb {M})/K(C))$ be an automorphism with $\sigma (K(A)) = K'$ . Let $\overline {c}$ be a basis of $V(C)$ . Let $C' = \langle K(A), C \rangle $ , so $V(C')$ is the extension of scalars of $V(C)$ to $K(A)$ . Let $\overline {a}$ be a $K(A)$ -basis of $V(A)$ over $V(C')$ ( $\overline {a}$ is possibly the empty tuple), so $\overline {a}\overline {c}$ is a basis of $V(A)$ as a $K(A)$ -vector space. Let $A'$ be the $2$ -sorted LLA over $K'$ with basis $\overline {a}\overline {c}$ , viewed naturally as an extension of $C" = \langle K',C \rangle $ , with structure induced from A by the isomorphism $\sigma $ , i.e., define $\tilde {\sigma } : A \to A'$ by

$$ \begin{align*}\sum \alpha_{i} a_{i} + \sum \beta_{j} c_{j} \mapsto \sum \sigma(\alpha_{i}) a_{i} + \sum \sigma(\beta_{j}) c_{j}, \end{align*} $$

and we define an LLA structure on $A'$ so that $\tilde {\sigma }$ is an isomorphism. By quantifier elimination, we can embed $A'$ into $\mathbb {M}$ over $C"$ . Let $A"$ denote the image and let $\overline {a}"$ denote the image of $\overline {a}$ . Note that $K(A") = K'$ .

Now let $\tilde {K} = K'K(B) = K(A")K(B)$ and define $\tilde {A}" = \langle \tilde {K},A" \rangle $ , $\tilde {B} = \langle \tilde {K},B \rangle $ , and $\tilde {C} = \langle \tilde {K},C \rangle = \langle \tilde {K},C' \rangle $ . By Fact 2.6, there is some $\tilde {D}$ , isomorphic over $\tilde {C}$ to $\tilde {A}"$ as an LLA over $\tilde {K}$ such that . By quantifier elimination, we may assume $\tilde {D}$ is embedded in $\mathbb {M}$ over $\tilde {B}$ and we have $\tilde {D} \equiv _{\tilde {C}} \tilde {A}"$ . Let $\overline {d}$ be the tuple corresponding to $\overline {a}"$ in $\tilde {D}$ . Then let $D = \langle K', \overline {d},\overline {c} \rangle $ . Note that, by construction, $D \equiv _{C} A$ and $\langle \tilde {K},D \rangle = \tilde {D}$ . Moreover, we have

$$ \begin{align*}\tilde{K} = K(D)K(B) \subseteq K(\langle D,B \rangle) \subseteq K(\langle \tilde{D},\tilde{B} \rangle) = \tilde{K}, \end{align*} $$

so $K(\langle D,B \rangle ) = \tilde {K}$ and we have as desired.

Finally, we restrict to the case when the field is algebraically closed and prove stationarity. Suppose A, $B_{0}$ , $B_{1}$ are algebraically closed sets containing an algebraically closed set C. Suppose $B_{0} \equiv _{C} B_{1}$ and for $i = 0,1$ . Let $g: B_{0} \to B_{1}$ be a C-isomorphism witnessing $B_{0} \equiv _{C} B_{1}$ . Define $\tilde {K}_{0} = K(\langle A,B_{0} \rangle )$ and $\tilde {K}_{1} = K(\langle A,B_{1} \rangle )$ . By stationarity of forking independence in ACF and the fact that $\tilde {K}_{0} = K(A)K(B_{0})$ and $\tilde {K}_{1} = K(A)K(B_{1})$ , we know that $g|_{K(B_{0})}$ extends to a $K(A)$ -isomorphism $f:\tilde {K}_{0} \to \tilde {K}_{1}$ . Since $\langle \tilde {K}_{i} \rangle = \tilde {K}_{i}$ (i.e., $V(\langle \tilde {K}_{i} \rangle ) = 0$ ) for $i = 0,1$ , we know by quantifier elimination that f extends to an automorphism $\sigma \in \mathrm {Aut}(\mathbb {M}/K(A))$ .

For $i = 0,1$ , let $\tilde {A}_{i} = \langle \tilde {K}_{i},A \rangle $ , $\tilde {B}_{i} = \langle \tilde {K}_{i},B_{i} \rangle $ , and $\tilde {C}_{i} = \langle \tilde {K}_{i},C \rangle $ . Since $V(\tilde {A}_{i}) \cong V(A) \otimes _{K(A)} \tilde {K}_{i}$ for $i =0,1$ , we have an isomorphism $\tilde {f}: \tilde {A}_{0} \to \tilde {A}_{1}$ induced by the isomorphism $v \otimes c \mapsto v \otimes f(c)$ for $v \in V(A)$ and $c \in \tilde {K}_{0}$ . Similarly, we have a map $\tilde {g}: \tilde {B}_{0} \to \tilde {B}_{1}$ induced by the isomorphism $V(B_{0}) \otimes _{K(B_{0})} \tilde {K}_{0} \to V(B_{1}) \otimes _{K(B_{1})} \tilde {K}_{1}$ given by $v \otimes c \mapsto g(v) \otimes f(c)$ for $v \in V(B_{0})$ and $c \in \tilde {K}_{0}$ . By construction, $\tilde {f}|_{\tilde {C}_{0}} = \tilde {g}|_{\tilde {C}_{0}}$ . Let $L = \sigma ^{-1}(\tilde {A}_{1} \otimes _{\tilde {C}_{1}} \tilde {B}_{1})$ , which may be regarded as an LLA over $\tilde {K}_{0}$ .

The maps $\sigma ^{-1} \circ \tilde {f}$ and $\sigma ^{-1} \circ \tilde {g}$ may be regarded as homomorphisms of LLAs over $\tilde {K}_{0}$ , which agree on $\tilde {C}_{0}$ , and therefore, by the universal property of the free amalgam, there is a unique map $h: \tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0} \to L$ extending $\sigma ^{-1} \circ \tilde {f}$ and $\sigma ^{-1} \circ \tilde {g}$ (note that it makes sense to say that this map extends $\sigma ^{-1} \circ \tilde {f}$ and $\sigma ^{-1} \circ \tilde {g}$ because we are identifying $\tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0}$ with $\langle \tilde {A}_{0},\tilde {B}_{0} \rangle $ ). By invariance, L is the free amalgam of $\sigma ^{-1}(\tilde {A}_{1})$ and $\sigma ^{-1}(\tilde {B}_{1})$ over $\sigma ^{-1}(\tilde {C}_{1})$ . The same argument, applied to the maps $\tilde {f}^{-1} \circ \sigma : \sigma ^{-1}(\tilde {A}_{1}) \to (\tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0})$ and $\tilde {g}^{-1} \circ \sigma : \sigma ^{-1}(\tilde {B}_{1}) \to (\tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0})$ implies that there is a map $L \to (\tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0})$ extending $\tilde {f}^{-1} \circ \sigma $ and $\tilde {g}^{-1} \circ \sigma $ , which must therefore be the inverse of h. This shows h is an isomorphism.

It follows, then, that $\sigma \circ h : \tilde {A}_{0} \otimes _{\tilde {C}_{0}} \tilde {B}_{0} \to \tilde {A}_{1} \otimes _{\tilde {C}_{1}} \tilde {B}_{1}$ is an isomorphism. Since $\sigma \circ h$ extends $\tilde {f}$ , which was defined to fix both $V(A)$ and $K(A)$ , we know that $\sigma \circ h$ fixes A. Additionally, as $\sigma \circ h$ extends $\tilde {g}$ , it takes $B_{0}$ to $B_{1}$ . By quantifier elimination, this shows $B_{0} \equiv _{A} B_{1}$ , which proves stationarity.

Proof. Our assumption gives us that, in the preceding proof, we may take ${K' = K(A)}$ and $\sigma $ to be the identity. Then the D constructed there will have $D \equiv _{CK(A)} A$ .

Proof. Let M be a model of $T^+$ and assume a, b, c, and d are finite tuples satifying the hypotheses of the statement of “weak transitivity”. Let $C = \mathrm {acl}(Md)$ , $A = \mathrm {acl}(aC)$ , and $B = \mathrm {acl}(bC)$ . Additionally, set $A_{0} = \mathrm {acl}(aM)$ and $B_{0} = \mathrm {acl}(bM)$ .

Our assumptions imply and . Therefore, by transitivity and monotonicity, we get the desired .

The proofs of [Reference d’Elbée, Müller, Ramsey and Siniora10, Lemma 5.11] and [Reference d’Elbée, Müller, Ramsey and Siniora10, Corollary 5.12] readily adapt to the two-sorted case, though we give details on one point. The claim of [Reference d’Elbée, Müller, Ramsey and Siniora10, Lemma 5.11] is that if, given LLAs $E,F,G$ over a fixed field with $G \subseteq E,F$ , then, assuming E and F are either heir or coheir independent over G, then it follows that if $e_{1}, \ldots , e_{n} \in E$ and $\langle F, e_{<i}\rangle $ is an ideal of $\langle F, e_{\leq i} \rangle $ for all $i \leq n$ , then $\langle G, e_{<i}\rangle $ is an ideal of $\langle G,e_{\leq i} \rangle $ for all $i \leq n$ . We claim this follows the weaker assumption that E and F are algebraically independent over G. To see this, assume E and F are algebraically independent over G and suppose we are given $e_{1}, \ldots , e_{n} \in E$ such that $\langle F, e_{<i}\rangle $ is an ideal of $\langle F, e_{\leq i} \rangle $ for all $i \leq n$ . Assume for induction that we have shown for some $m < n$ that $\langle G, e_{<i}\rangle $ is an ideal of $\langle G,e_{\leq i} \rangle $ for all $i \leq m$ . For the induction step, we must show $[e_{m+1},v] \in \langle G, e_{\leq m}\rangle $ for $v \in \langle G, e_{\leq m} \rangle $ . Write $v = g + \sum _{i = 1}^{m} \lambda _{i} e_{i}$ . By our hypothesis, $[e_{m+1}, v] \in \langle F, e_{\leq m} \rangle $ so we may write

$$ \begin{align*}[e_{m+1},v] = f + \sum_{i = 1}^{m} \mu_{i}e_{i}, \end{align*} $$

for some $f \in F$ . But then $f = [e_{m+1},v] - \sum _{i=1}^{m} \mu _{i}e_{i}\in E \cap F = G$ , by our assumption of algebraic independence, so we have $[e_{m+1},v] \in \langle G, e_{\leq m} \rangle $ as desired.

The argument of [Reference d’Elbée, Müller, Ramsey and Siniora10, Theorem 5.13], then, gives us that where $\tilde {A}^{\prime }_{0} = \langle K(\langle A,B \rangle ),A \rangle $ , $\tilde {B}_{0}^{\prime } = \langle K(\langle A,B \rangle ), B \rangle $ , and $\tilde {M}' = \langle K(\langle A,B \rangle ), M \rangle $ . We claim that this implies . We are left to show that if $\tilde {A}_{0} = \langle K(\langle A_{0},B_{0} \rangle ),A\rangle $ , $\tilde {B}_{0} = \langle K(\langle A_{0},B_{0}),B_{0}\rangle $ , and $\tilde {M} = \langle K(\langle A_{0},B_{0} \rangle ),M\rangle $ , then , and this follows immediately by Lemma 4.17.

Proof. First, we use NSOP $_{1}$ in the field sort to find some $K_{*}$ such that

$$ \begin{align*}K_{*} {\vDash} \mathrm{tp}_{L^{\dagger}}(K(C_{0})/K(A)) \cup \mathrm{tp}_{L^{\dagger}}(K(C_{1})/K(B)). \end{align*} $$

By Lemma 4.14, we may choose $K_{*}$ so that . By quantifier elimination relative to the field sort (see Corollary 3.3 or $(\star )$ above the latter), choose some $C_{*} \equiv _{M} C_{0} \equiv _{M} C_{1}$ so that $K(C_{*}) = K_{*}$ . Notice that we have $C_{*} \equiv _{MK(A)} C_{0}$ and $C_{*} \equiv _{MK(B)} C_{1}$ , by Lemma 3.5.

Choose some $\sigma _{0} \in \mathrm {Aut}(\mathbb {M}/MK(A))$ with $\sigma _{0}(C_{0}) = C_{*}$ and set $K_{0} = \sigma _{0}(K(\langle A,C_{0} \rangle ))$ . Likewise, choose some $\sigma _{1} \in \mathrm {Aut}(\mathbb {M}/MK(B))$ with $\sigma _{1}(C_{1}) = C_{*}$ and $K_{1} = \sigma _{1}(K(\langle B,C_{1} \rangle ))$ . Recall that we have already arranged that . By extension and symmetry, we may choose $K_{0}^{\prime } \equiv _{K(A)K_{*}} K_{0}$ and $K^{\prime }_{1} \equiv _{K(B)K_{*}} K_{1}$ such that . Necessarily, we have also, then, $K_{0}^{\prime } \equiv _{K(A)C_{*}} K_{0}$ and $K^{\prime }_{1} \equiv _{K(B)C_{*}} K_{1}$ since the field sort is stably embedded. So without loss of generality, $K_{0} = K_{0}^{\prime }$ and $K_{1} = K_{1}^{\prime }$ .

Notice that, since we have $K(\langle A,C_{0} \rangle )C_{0} \equiv _{MK(A)} K_{0}C_{*}$ and $K(\langle B,C_{1} \rangle )C_{1} \equiv _{MK(B)} K_{1}C_{*}$ , we can find $A'$ and $B'$ such that

$$ \begin{align*}K(\langle A,C_{0} \rangle) A C_{0} \equiv_{MK(A)} K_{0}A'C_{*} \end{align*} $$

and

$$ \begin{align*}K(\langle B,C_{1} \rangle) BC_{1} \equiv_{MK(B)} K_{1}B'C_{*}. \end{align*} $$

By construction, $K(\langle A',C_{*} \rangle ) = K_{0}$ and $K(\langle B',C_{*} \rangle ) = K_{1}$ . Thus, applying Lemma 4.20, we may assume .

Note that we have and , where the tilde indicates the extension of scalars up to $\tilde {K} = K_{0}K_{1}$ . This follows from the fact that linear independence is preserved under extension of scalars. Hence, by Lemma 4.22, we have again as LLAs over $\tilde {K}$ . From this it follows, by Lemma 4.17, that $\langle A',B' \rangle $ is the free amalgam of $A"$ and $B"$ over $M"$ , where $A"$ , $B"$ , and $M"$ are LLAs over $K(\langle A',B' \rangle ) = K(A')K(B') = K(A)K(B)$ obtained from $A'$ , $B'$ , and M by extension of scalars.

But, as $A \equiv _{M} A'$ and $K(A) = K(A')$ , we have, by Lemma 3.5 that $A \equiv _{MK(A)} A'$ and therefore there is an isomorphism f over $MK(A)$ from A to $A'$ . By extending scalars to $K(A)K(B)$ , f lifts uniquely to an $K(A)K(B)$ -isomorphism $f'$ from $A"$ to the extension of scalars of $A"$ to an LLA over $K(A)K(B)$ . We may regard $f"$ as an embedding of LLAs over $\langle M,K(A),K(B)\rangle $ from $A"$ to $\langle A',B' \rangle $ , with $f(A) = f(A')$ . Arguing similarly, we can find some $K(A)K(B)$ -embedding $g: B" \to \langle A',B'\rangle $ of LLAs over $\langle M,K(A),K(B)\rangle $ with $g(B) = B'$ . By the universal property for the free amalgam, there is a unique $K(A)K(B)$ -embedding of LLAs $h : \langle A,B \rangle \to \langle A',B' \rangle $ over M extending f and g. By a symmetric argument applied to $f^{-1}$ and $g^{-1}$ , we see that h must be an isomorphism. In other words, we have shown there is some $\sigma \in \mathrm {Aut}(\mathbb {M}/MK(A)K(B))$ with $\sigma (A'B') = AB$ . Let $C_{**} = \sigma (C_{*})$ .

Note that we have $A'C_{*} \equiv _{M} AC_{**}$ and $B'C_{*} \equiv _{M} BC_{*}$ , hence

$$ \begin{align*}C_{**} {\vDash} \mathrm{tp}(C_{0}/A) \cup \mathrm{tp}(C_{1}/B) \end{align*} $$

as desired.

Proof. By Fact 4.8, it suffices to show that if $M {\vDash } T^{+}$ and $(A_{i})_{i < \omega }$ is a coheir sequence over M consisting of models of T that contain M then, setting $p(X,Y) = \mathrm {tp}(A_{0},A_{1}/M)$ , we have

$$ \begin{align*}p(X_{0},X_{1}) \cup p(X_{1},X_{2}) \cup p(X_{2},X_{3}) \cup p(X_{3},X_{0}) \end{align*} $$

is consistent.

Choose $A_{2}^{\prime } \equiv _{A_{1}} A_{2}$ with . By weak transitivity, we have .

Pick $A^{\prime }_{1}$ such that

$$ \begin{align*}A^{\prime}_{1}A_{2}^{\prime} \equiv_{M} A_{2}A_{1} \left( \equiv_{M} A^{\prime}_{2}A_{1} \right). \end{align*} $$

Note that we have ${\vDash } p(A_{2}^{\prime },A^{\prime }_{1})$ and .

Likewise, choose $A^{\prime \prime }_{1}$ such that

$$ \begin{align*} A^{\prime\prime}_{1}A_{0} \equiv_{M} A_{0}A_{1}. \end{align*} $$

Then again we have ${\vDash } p(A^{\prime \prime }_{1},A_{0})$ and . Since $A^{\prime }_{1} \equiv _{M} A^{\prime \prime }_{1}$ , we may apply Proposition 4.23 to obtain some $A_{*}$ such that

$$ \begin{align*}A_{*} {\vDash} \mathrm{tp}(A^{\prime}_{1}/A^{\prime}_{2}) \cup \mathrm{tp}(A^{\prime\prime}_{1}/A_{0}). \end{align*} $$

Therefore, we have

$$ \begin{align*}{\vDash} p(A_{0},A_{1}) \cup p(A_{1},A^{\prime}_{2}) \cup p(A^{\prime}_{2},A_{*}) \cup p(A_{*},A_{0}), \end{align*} $$

completing the proof.

Question 4.25. What is Conant-independence in T? We conjecture that it is field independence plus vector space independence for the algebraic closures—that is, we conjecture Conant-independence coincides with defined above, over models.

Proof. Let $(a_i)_{i<\omega }$ be an indiscernible sequence and for $p(x,y) = \mathrm {tp}(a_0,a_1)$ , we need to prove that

$$\begin{align*}p(x_0,x_1)\cup p(x_1,x_2)\cup p(x_2,x_3)\cup p(x_3,x_0)\end{align*}$$

is consistent. By Fact 4.8, there is a model M such that for all i. By full existence, there exists $a_2^{*}\equiv _{Ma_1} a_2$ such that . From and we get by invariance and by definition. Using weak transitivity, we conclude . As $a_0\equiv _M a_1$ there exists $c_1$ such that $c_1a_0\equiv _M a_0a_1$ and by invariance, . Similarly, as $a_2^{*}\equiv _M a_1$ there exists $c_2$ such that $a_1a_2^{*}\equiv _M a_2^{*}c_2$ and by invariance . We have $c_1\equiv _M c_2$ , , hence by the weak independence theorem over models, we conclude that there exists $a_3^{*}$ such that $a_3^{*}a_0\equiv _M c_1a_0\equiv _M a_0a_1$ and $a_3^{*}a_2^{*}\equiv _M c_2a_2^{*}\equiv _M a_2^{*}a_1\equiv _M a_2a_1$ . As $a_0a_1\equiv a_1a_2$ , we conclude that

$$\begin{align*}a_0a_1\equiv a_1a_2^{*}\equiv a_2^{*} a_3^{*}\equiv a_3^{*}a_0\end{align*}$$

hence, the type above is consistent.

Question 4.27. It would be interesting to prove that $T^{+}$ is NSOP $_{4}$ in cases where the theory of the field is not assumed to be NSOP $_{1}$ . Some precise variants of this question.

1. (1) Suppose $T^{\dagger }$ is NSOP $_{4}$ . Does it follow that $T^{+}$ is NSOP $_{4}$ ?

2. (2) Suppose $T^{\dagger }$ has symmetric Conant-independence. Does it follow that $T^{+}$ is NSOP $_{4}$ ?

3. (3) Suppose $T^{\dagger }$ is a theory of curve-excluding fields. Does it follow that $T^{+}$ is NSOP $_{4}$ ?