Fact 2.1 (Axiomatisation of $\mathrm {ACFG}$ )

The theory $\mathrm {ACFG}$ is axiomatised by adding to $\mathrm {ACF}_{G}$ the following $\mathscr {L}_G$ -sentences, for all tuples of variables $x'\subset x$ , $y' \subset y$ and $\mathscr {L}_{\mathrm {ring}}$ -formulae $\phi (x,y)$

$$ \begin{align*}\forall y (\langle {y'} \rangle\cap G = \{ {0} \} \wedge \theta_{\phi}(y)) \rightarrow (\exists x \phi(x,y)\wedge \langle {xy'} \rangle\cap G = \langle {x'} \rangle).\end{align*} $$

Proof Apart from Transitivity and Local Character, all properties have been proven in [Reference d’Elbée16, Theorem 4.2, Example 5.21].

Transitivity. Assume that and . We may assume that $A = \overline {ABC},B=\overline {CB}$ and $D=\overline {CD}$ . By Monotonicity, it is sufficient to show that . We clearly have by Transitivity of . We show that $G(A+D) = G(A)+G(D)$ . By we have $G(A + \overline {BD}) = G(A)+G(\overline {BD})$ . It follows that $G(A+D)$ is included in $(A+D)\cap (G(A)+G(\overline {BD}))$ , which, by modularity, is equal to

$$ \begin{align*}G(A)+ ((A+D)\cap G(\overline{BD})) = G(A)+ G(A\cap \overline{BD} + D).\end{align*} $$

As , $A\cap \overline {BD} = B$ . By , $G(B+D) = G(B)+G(D)$ hence

$$ \begin{align*}G(A+D) = G(A)+G(B)+G(D) = G(A)+G(D).\end{align*} $$

Local Character. We start with a claim.

Claim. Let $A,B$ be subsets of $(K,G)$ with B a subgroup of $(K,+)$ , then there exists $C\subseteq B$ with $\lvert {C} \rvert \leq \lvert {A} \rvert $ such that

$$ \begin{align*}G(A+B) = G(A+C)+G(B).\end{align*} $$

Proof of the claim

For each $a\in A$ define $C(a)$ to be the set of those $c\in B$ such that $a+c\in G$ . Take $c(a)\in C(a)$ for each a such that $C(a)$ is nonempty, and set

$$ \begin{align*}C = \{ {c(a) \mid a\in A \text{ and }C(a)\neq \emptyset} \}.\end{align*} $$

Now if $g\in G(A+B)$ then $g=a+b$ with $a\in A$ , $b\in B$ . We have $C(a)$ nonempty so we can write for $c = c(a)$

$$ \begin{align*}g =(a+ c)+ (b-c).\end{align*} $$

It follows that $b-c \in G(B)$ hence $g\in G(A+C)+G(B)$ . The reverse inclusion is trivial. ⊣

Let a be a finite tuple and B an algebraically closed set. We construct two sequences $(A_i)_{i<\omega }$ and $(D_i)_{i<\omega }$ such that the following holds for all $n<\omega $ :

1. (1) $A_n\subseteq A_{n+1}\subseteq \overline {Ba}$ and $D_n\subseteq A_{n}$ .

2. (2) $G(A_n + B)\subseteq G(A_{n+1})+G(B)$ .

3. (3) .

4. (4) $\lvert {A_n} \rvert \leq \aleph _0$ .

Using Local Character for there exists a countable set $D_0\subseteq B$ such that . We define $A_0 = \overline {a D_0}$ . Assume that $D_n$ and $A_n$ has been constructed and that $\lvert {A_n} \rvert \leq \aleph _0$ . By the claim there exists $C\subseteq B$ with $\lvert {C} \rvert \leq \aleph _0$ such that $G(A_n+B) = G(A_n+C)+G(B)$ . Using Local Character Footnote ¹ of on the set $A_nC$ there exists $D_{n+1}\subseteq B$ with $\lvert {D_{n+1}} \rvert \leq \aleph _0$ such that . We set $A_{n+1} = \overline {A_n C D_{n+1}}$ . Note that $A_n+C \subseteq A_{n+1}$ so $G(A_n+B)\subseteq G(A_{n+1})+G(B)$ .

Now set $A_\omega = \bigcup _{i<\omega } A_i$ and $D_{\omega } = \bigcup _{i<\omega } D_i$ . We have $\lvert {A_\omega } \rvert \leq \aleph _0$ and $\lvert {D_\omega } \rvert \leq \aleph _0$ . We claim that

If u is a finite tuple from $A_\omega $ , then $u\subseteq A_{n}$ for some n, so as

we have

. Now as $D_\omega \subseteq B$ , we use Base Monotonicity of

to conclude that

. As this holds for every finite tuple u from $A_\omega $ , we conclude that

It remains to show that $G(A_\omega +B) = G(A_\omega )+G(B)$ . If $g\in G(A_\omega +B)$ then there is some n such that $g\in A_n+B$ and so

$$ \begin{align*}g\in G(A_n+B)\subseteq G(A_{n+1})+G(B)\subseteq G(A_\omega)+G(B).\end{align*} $$

The reverse inclusion being trivial, we conclude that $G(A_\omega + B)= G(A_\omega )+G(B)$ , so

. As $a\subseteq A_\omega $ we conclude by Monotonicity of

.⊣

Proof The first assertion holds because satisfies -amalgamation over algebraically closed sets (Theorem 2.4). The proof is a classical induction similar to the proof of Lemma 4.12 or [Reference Conant and Kruckman15, Proposition 4.11].

(1) implies (2) is a particular case of the first assertion. (2) implies (3) follows from the fact that such sequence exists, which follows from Full Existence of . We show that (3) implies (1). Assume that and let $\Lambda (x,b,c)$ be as in [Reference d’Elbée16, Lemma 3.6]. If (3) holds, then in particular $\{ {\Lambda (x,b_i,c)\mid i<\omega } \}$ is consistent, for some $(b_i)_{i<\omega }$ such that $b_i\equiv _C b$ and . This contradicts [Reference d’Elbée16, Lemma 3.6].⊣

Proof (1) implies (2) follows from Proposition 2.6, and (2) implies (3) holds since satisfies Full Existence. Assume that (3) holds for some $a'$ and indiscernible sequence $(b_i)_{i<\omega }$ such that for all $i<\omega $ . In particular, $(b_i)_{i<\omega }$ is a Morley sequence in the sense of $\mathrm {ACF}_p$ , and $a'b_i \equiv ^{\mathrm {ACF}}_C ab$ for all $i<\omega $ . As is forking independence in the sense of $\mathrm {ACF}_p$ , we have . By Proposition 2.6 we have .⊣

Proof This is [Reference d’Elbée16, Lemma 3.10].⊣

Proof Let $I\subset \{ {1,\dots ,n} \}$ . As P is non- $\mathbb {F}_p$ -flat, there exists a zero t of P in an extension of K such that no non-nontrivial $\mathbb {F}_p$ -linear combination of t falls in K. It follows that $(\overline {K(t)}, G+\langle {t_i \mid i\in I} \rangle )$ is an $\mathscr {L}_G$ -extension of $(K,G)$ . Indeed $(G+\langle {t_i \mid i\in I} \rangle )\cap K = G$ . Furthermore $t_j\in (G+\langle {t_i \mid i\in I} \rangle )$ if and only if $j\in I$ . As $(K,G)$ is existentially closed in $(\overline {K(t)}, G+\langle {t_i \mid i\in I} \rangle )$ , we have that

$$ \begin{align*}(K,G)\models \exists x (P(x)=0 \wedge \bigwedge_{i\in I} x_i\in G \wedge \bigwedge_{j\notin I} x_j\notin G).\\[-44pt]\end{align*} $$

⊣

Proof Assume that P is $\mathbb {F}_p$ -flat over K. If $\lvert {X} \rvert = 1$ , then P satisfies the conclusion. Assume that $\lvert {X} \rvert>1$ . Let $t_2,\dots ,t_n$ be algebraically independent over K, and consider $P(X_1,t_2,\dots ,t_n)$ . This polynomial has zeros in $\overline {K(t_2,\dots ,t_n)}$ hence by $\mathbb {F}_p$ -flatness each root u satisfies $\lambda _1u + \lambda _2t_2+\dots +\lambda _n t_n =b$ for some non-zero tuple $\lambda _1,\dots ,\lambda _n$ from $\mathbb {F}_p$ and $b\in K$ . By hypothesis on $t_2,\dots ,t_n$ we have that $\lambda _1\neq 0$ . It follows that $X_1-\lambda _1^{-1}(\lambda _2t_2+\dots +\lambda _n t_n -b)$ divides $P(X_1,t_2,\dots ,t_n)$ hence $\lambda _1X_1+\dots +\lambda _n X_n -b$ divides P, as $K[X_1,t_2,\dots ,t_n]\cong K[X]$ . If $\lambda _i = 0$ for some i, then the tuple $(0,\dots ,t,\dots ,0)$ with t transcendental over K at the i-th coordinate, is a zero of P that contradicts the $\mathbb {F}_p$ -flatness. It follows that P is of the desired form. The other direction is trivial.⊣

Proof (1) For all $b\in K$ the polynomial $XY-b$ is not $\mathbb {F}_p$ -flat by Lemma 2.11, so we conclude using Lemma 2.10.

(2) From (1), every element in K is product of two elements in G, so any $\mathscr {L}_G$ -formula $\phi (x,a_1,\dots ,a_n)$ is equivalent to $\phi (x,g_1h_1,\dots ,g_nh_n)$ with $g_i,h_i\in G$ .

(3) For all $P\in K[X]\setminus (K+\mathbb {F}_p\cdot X)$ , $b\in K$ , the polynomial $Y+P(X)-b$ is not $\mathbb {F}_p$ -flat, similarly to (1).

Proof Using the first isomorphism theorem, it is sufficient to prove that the function $\zeta :K\rightarrow K/\zeta _1^{-1}(G) \times \dots \times K/\zeta _n^{-1}(G)$ defined by $\zeta (u) = (u+\zeta _1^{-1}(G),\dots , u+\zeta _n^{-1}(G))$ is onto. Let $c_1,\dots ,c_n\in K$ , we want to show that there exists $c\in K$ such that for all $i \zeta _i(c-c_i)\in G$ . Let t be a transcendental element over K, by model completeness of $\mathrm {ACF}_p$ , $\zeta _1,\dots ,\zeta _n$ are $\mathbb {F}_p$ -linearly independent definable endomorphisms of $(\overline {Kt},+)$ . Consider the $\mathscr {L}_G$ -structure

$$ \begin{align*}(\overline{Kt}, G+\langle {\zeta_i(t-c_i)\mid i\leq n} \rangle).\end{align*} $$

We have $(G+\langle {\zeta _i(t-c_i)\mid i\leq n} \rangle )\cap K = G+\langle {\zeta _i(t-c_i)\mid i\leq n} \rangle \cap K$ . For $\lambda _1,\dots ,\lambda _n\in \mathbb {F}_p$ , if $\sum _i \lambda _i \zeta _i(t-c_i)\in K$ then $\sum _i \lambda _i \zeta _i(t)\in K$ . It is standard that a definable additive endomorphism is of the form $x\mapsto a_1\mathrm {Frob}^{n_1}(x)+\cdots +a_k\mathrm {Frob}^{n_k}(x)$ with $n_i\in \mathbb {Z}$ (see [Reference Bouscaren, Delon, Hindry, Hrushovski, Lascar, Marker, Pillay, Wood and Ziegler6, Chapter 4, Corollary 1.5] and [Reference Humphreys21, Lemma A, VII, 20.3]) hence there is some k such that $t\mapsto (\sum _i \lambda _i \zeta _i(t))^{p^k}$ is polynomial. As t is transcendental over K, $(\sum _i \lambda _i \zeta _i)^{p^k} = 0$ , so $\sum _i \lambda _i \zeta _i = 0$ . As $\zeta _1,\dots ,\zeta _n$ are $\mathbb {F}_p$ -linearly independent, $\lambda _1=\dots =\lambda _n = 0$ . It follows that $(G+\langle {\zeta _i(t-c_i)\mid i\leq n} \rangle )\cap K = G$ , so $(\overline {Kt}, G+\langle {\zeta _i(t-c_i)\mid i\leq n} \rangle )$ extends $(K,G)$ . As $(K,G)$ is existentially closed in $(\overline {Kt}, G+\langle {\zeta _i(t-c_i)\mid i\leq n} \rangle )$ we have that $(K,G)\models \exists x \bigwedge _i \zeta _i(x-c_i)\in G$ , hence $\zeta $ is onto.⊣

Proof Assume that $\mathbb {F}_{p^n}\models \theta _\phi (b)$ . Then as $\theta _\phi $ is quantifier free, $\overline {{\mathbb {F}_p}}\models \theta _\phi (b)$ . It follows that for some elementary extension K of $\overline {{\mathbb {F}_p}}$ , there is some realisation a of $\phi (x,b)$ which is $\mathbb {F}_p$ -linearly independent over $\overline {{\mathbb {F}_p}}$ . In particular, in K, no nontrivial $\mathbb {F}_p$ -linear combination of a is a root of $X^{p^n}-X$ . As $\overline {\mathbb {F}_p} \equiv ^{\mathrm {ACF}} K$ , the following sentence holds in $\overline {\mathbb {F}_p}$ :

$$ \begin{align*}&\forall y (\theta_\phi(y)\\&\quad\rightarrow (\exists x \phi(x,y) \wedge \mbox{``no nontrivial}\ \mathbb{F}_p\mbox{-linear combination of} x \mbox{is a root of}\ X^{p^n}-X")).\end{align*} $$

Hence for some $m>n$ there exists a tuple a from $\mathbb {F}_{p^m}$ such that

$$ \begin{align*}\overline{\mathbb{F}_{p}}\models \phi(a,b)\wedge\ ``a\ \text{is}\ \mathbb{F}_p\text{-linearly independent over}\ \mathbb{F}_{p^n}".\end{align*} $$

As $\phi (x,y)$ is quantifier-free, we also have that

$$ \begin{align*}\mathbb{F}_{p^m}\models \phi(a,b)\wedge\ ``a\ \text{is}\ \mathbb{F}_p\text{-linearly independent over}\ \mathbb{F}_{p^n}".\\[-30pt]\end{align*} $$

⊣

Proof Start with the following claim.

From Proposition 2.1, the axioms for $\mathrm {ACFG}$ are given by the following scheme: for all quantifier free $\mathscr {L}_{\mathrm {ring}}$ -formula $\phi (x,y)$ , for all $0\leq k\leq \lvert {x} \rvert $ and $0\leq k'\leq \lvert {y} \rvert $

$$ \begin{align*}\forall y ( (\theta_\phi(y)\kern1.2pt{\wedge}\kern1.2pt \langle {y_1,\dots,y_{k'}} \rangle\cap G \kern1.2pt{=}\kern1.2pt \{ {0} \})\kern1.2pt{\rightarrow}\kern1.2pt (\exists x \phi(x,y)\kern1.2pt{\wedge}\kern1.2pt \langle {x,y_1,\dots,y_{k'}} \rangle\kern1.2pt{\cap}\kern1.2pt G \kern1.2pt{=}\kern1.2pt \langle {x_1,\dots,x_k} \rangle))\end{align*} $$

with the following convention: $a_1,\dots ,a_0 = \emptyset $ . We will denote the previous sentence by $\Gamma (\phi ,k,k')$ . Now we construct by induction a model of $\mathrm {ACFG}$ starting from $(\mathbb {F}_{p^n}, G_0)$ . Let $(\phi _i(x^i,y^i))_{i<\omega }$ be an enumeration of all quantifier-free formulae in $\mathscr {L}_{\mathrm {ring}}$ . We construct an increasing sequence $(n_j)_{j<\omega }$ starting with $n_0 = n$ and additive subgroups $G_j$ of $\mathbb {F}_{p^{n_j}}$ such that for all $s<\omega $ , for $\phi _1(x^1,y^1),\dots ,\phi _s(x^s,y^s)$ , for all $1\leq l \leq s$ , for all $0\leq k\leq \lvert {x^l} \rvert $ and $0\leq k'\leq \lvert {y^l} \rvert $ the following holds for all $\lvert {y^l} \rvert $ -tuples b from $\mathbb {F}_{p^{n_s}}\!$ :

If $\mathbb {F}_{p^{n_s}}\models \theta _{\phi _l}(b)\wedge \langle {b_1,\dots , b_{k'}} \rangle \cap G_s = \{ {0} \}$ then there exists $a^{l,k}$ an $\lvert {x^l} \rvert $ -tuple from $\mathbb {F}_{p^{n_{s+1}}}$ such that $\mathbb {F}_{p^{n_{s+1}}} \models \phi (a^{l,k},b)\wedge \langle {a^{l,k},b_1,\dots ,b_{k'}} \rangle \cap G_{s+1} = \langle {a^{l,k}_1,\dots ,a^{l,k}_k} \rangle $ . ( $\star $ )

Assume that for some $s<\omega $ we have $n_0,\dots ,n_s$ and $G_0\subseteq \mathbb {F}_{p^{n_0}},\dots , G_s\subseteq \mathbb {F}_{p^{n_s}}$ constructed as above. For every $i\kern1pt{\leq}\kern1pt s$ , we define as above $B_i= \{ {b\in \mathbb {F}_{p^{n_s}}^{\lvert {y^i} \rvert } \mid b\models \theta _{\phi _i}(y)} \}$ , and we apply the claim with $k_i = \lvert {x^i} \rvert +1$ , to get some $n_{s+1}>n_s$ . For each $1\leq i \leq s$ and $b\in B_i$ we have $\lvert {x^i} \rvert +1$ many $\lvert {x^i} \rvert $ -tuples $a^{i,1}(b), \dots , a^{i,k_i}(b)$ from $\mathbb {F}_{p^{n_{s+1}}}$ all $\mathbb {F}_p$ -independent over $\mathbb {F}_{p^{n_s}}$ and such that for all j, we have $\mathbb {F}_{p^{n_{s+1}}}\models \phi _i(a^{i,j}(b),b)$ . Now define $G_{s+1}$ to be

$$ \begin{align*}G_s\oplus \bigoplus_{1\leq i\leq s}\bigoplus_{b\in B_i}\langle {a_1^{i,2}(b)} \rangle\oplus\langle {a_1^{i, 3}(b),a_2^{i,3}(b)} \rangle\oplus \dots\oplus \langle {a_1^{i,k_i}(b),\dots,\oplus a_{k_i}^{i,k_i}(b)} \rangle.\end{align*} $$

In other words, $G_{s+1}$ is obtained by adding to $G_s$ the low triangle of the $(\lvert {x_i} \rvert +1)\times \lvert {x_i} \rvert $ matrix $(a^{i,j}_k(b))_{1\leq j\leq k_i,1\leq k\leq \lvert {x^i} \rvert }$ for each $i<s$ and $b\in B_i$ :

Now we have for each $1\leq i\leq s$ and any $0\leq k\leq \lvert {x^i} \rvert $ and $0\leq k'\leq \lvert {y^i} \rvert $ , if $b\in B_i$ , then there exists $a^{i,k}(b)\in \mathbb {F}_{p^{n_{s+1}}}^{\lvert {x^i} \rvert }$ such that $ \mathbb {F}_{p^{n_{s+1}}}\models \phi _i(a^{i,k}(b), b)$ . By construction and $\mathbb {F}_p$ -linear independence of all the $a^{i,k}$ , if $\langle {b_1,\dots ,b_{k'}} \rangle \cap G_s = \{ {0} \}$ then we have $\langle {a^{i,k},b_1,\dots ,b_{k'}} \rangle \cap G_{s+1} = \langle {a^{i,k}_1,\dots ,a^{i,k}_k} \rangle $ . By induction we construct a family $(\mathbb {F}_{p^{n_i}},G_i)$ satisfying ( $\star $ ). Now let

$$ \begin{align*}G= \bigcup_{i<\omega} G_i \subseteq \overline{\mathbb{F}_p}.\end{align*} $$

By construction, we have that $(\overline {\mathbb {F}_p},G)$ is a model of $\mathrm {ACFG}$ .⊣

Proof The density of $\mathcal {C}$ in $\mathrm {Sg}(\overline {\mathbb {F}_p})$ follows directly from Proposition 2.16: every clopen $\mathbb {B}(G_0, \mathbb {F}_{p^n})$ contains an element of $\mathcal {C}$ .

We show that $\mathcal {C}$ is a $G_\delta $ . First, from Proposition 2.1, $\mathrm {ACFG}$ is axiomatised by adding to the theory $\mathrm {ACF}_G$ the following $\mathscr {L}_G$ -sentences, for all tuples of variables $x'\subset x$ , $y' \subset y$ and $\mathscr {L}_{\mathrm {ring}}$ -formulae $\phi (x,y)$

$$ \begin{align*}\forall y (\langle {y'} \rangle\cap G = \{ {0} \} \wedge \theta_{\phi}(y)) \rightarrow (\exists x \phi(x,y)\wedge \langle {xy'} \rangle\cap G = \langle {x'} \rangle)\end{align*} $$

which is equivalent to

$$ \begin{align*}\forall y\exists x \left[\neg \theta_\phi(y))\vee \langle {y'} \rangle\cap G \neq \{ {0} \} \vee (\phi(x,y)\wedge \langle {xy'} \rangle\cap G = \langle {x'} \rangle)\right].\end{align*} $$

Let $\phi (x,y)$ , $x'\subseteq x$ and $y'\subseteq y$ be given. Let b be a $\lvert {y} \rvert $ -tuple, and consider the set

$$ \begin{align*}\mathscr{O}_b =\!\! \!\bigcup_{a\in \overline{\mathbb{F}_p}^{\lvert {x} \rvert},\overline{\mathbb{F}_p}\models \phi(a,b)}\!\!\! \{ {H\kern1.5pt{\in}\kern1.5pt \mathrm{Sg}(\overline{\mathbb{F}_p}) \kern1pt{\mid}\kern1pt \langle {b'} \rangle\cap H \kern1.5pt{\neq}\kern1.5pt \{ {0} \}} \}\kern1.5pt{\cup}\kern1.5pt \{ {H\kern1.5pt{\in}\kern1.5pt \mathrm{Sg}(\overline{\mathbb{F}_p}) \kern1pt{\mid}\kern1pt \langle {ab'} \rangle\cap H \kern1pt{=}\kern1pt \langle {a'} \rangle} \kern-1pt\}.\end{align*} $$

$\mathscr {O}_b$ depends on b and the choice of $\phi (x,y)$ , $x'\subseteq x$ and $y'\subseteq y$ . $\mathscr {O}_b$ is the set of $H\in \mathrm {Sg}(\overline {\mathbb {F}_p})$ such that there exists $a\in \overline {\mathbb {F}_p}$ such that $\overline {\mathbb {F}_p} \models \phi (a,b)$ and if $\langle {b'} \rangle \cap H = \{ {0} \}$ , then $\langle {ab'} \rangle \cap H = \langle {a'} \rangle $ .

The set $\{ {H\kern1pt{\neq}\kern1pt \mathrm {Sg}(\overline {\mathbb {F}_p}) \mid \langle {b'} \rangle \cap H \kern1pt{\neq}\kern1pt \{ {0} \}} \}$ is equal to $\bigcup _{u\in \langle {b'} \rangle \setminus \{ {0} \}}\{ {H\kern1pt{\in}\kern1pt \mathrm {Sg}(\overline {\mathbb {F}_p}) \mid u\kern1pt{\in}\kern1pt H} \}$ which is open since each set $\{ {H\in \mathrm {Sg}(\overline {\mathbb {F}_p}) \mid u\in H} \}$ is open. The set $\{ H\in \mathrm {Sg}(\overline {\mathbb {F}_p}) \mid \langle {ab'} \rangle \cap H = \langle {a'} \rangle \}$ is also open, so $\mathscr {O}_b$ is open. Now it is an easy checking that

$$ \begin{align*}\mathcal{C} = \bigcap_{\phi(x,y), x'\subseteq x, y'\subseteq y}\bigcap_{b\in \overline{\mathbb{F}_p}^{\lvert {y} \rvert},\overline{\mathbb{F}_p}\models \theta_\phi(b)} \mathscr{O}_b.\end{align*} $$

Hence $\mathcal {C}$ is a countable intersection of open sets, i.e., $\mathcal {C}$ is $G_\delta $ .⊣

Proof Let x be a tuple from an elementary extension of the field K, such that and such that $x \equiv ^{\mathrm {ACF}}_C a$ , and $f : \overline {Cx} \rightarrow \overline {Ca}$ a field isomorphism over C sending x to a. Let $G_{Cx} = f^{-1}(G(\overline {Ca}))$ . Consider now the subgroup of $\overline {CDbx}$ defined by

$$ \begin{align*}H = G_{Cx}+G(\overline{CbD}) + \langle {x_i - b_i \mid i\leq \lvert {x} \rvert} \rangle.\end{align*} $$

We show that the type in the sense of $\mathrm {ACFG}$ defined by the pair $(\overline {CDbx}, H)$ is consistent. As we have $\overline {CDbx}\cap K = \overline {CDb}$ . In order to prove that $H\cap \overline {CDb} = G(\overline {CDb})$ , it suffices to show that

$$ \begin{align*}\overline{CDb}\cap (G_{Cx}+ \langle {x_i - b_i \mid i\leq \lvert {x} \rvert} \rangle)\subseteq G(\overline{C}).\end{align*} $$

Assume that $g_{Cx}+\sum _i\lambda _i(x_i-b_i) \in \overline {CDb}$ , where $g_{Cx}\in G_{Cx}$ . It follows that $g_{Cx}+\sum _i\lambda _i x_i \in \overline {CDb}$ . On the other hand $g_{Cx}+\sum _i\lambda _i x_i \in \overline {Cx}$ . As we have $\overline {Cx}\cap \overline {CDb} = \overline {C}$ hence $g_{Cx}+\sum _i\lambda _i x_i \in \overline {C}$ . Apply $\pi \circ f$ to get that $\sum _i\lambda _i\pi (a_i)\in \pi (\overline {C})$ hence by hypothesis $\lambda _i = 0$ for all $i\leq \lvert {x} \rvert $ . It follows that $g_{Cx}\in \overline {C}$ and so $g_{Cx}\in G(\overline {C})$ . We have showed that $\overline {CDb}\cap (G_{Cx}+ \langle {x_i - b_i \mid i\leq \lvert {x} \rvert } \rangle )\subseteq G(\overline {C})$ . The type is consistent by §2.1, so realised by say $a'$ . As we have . In order to show that $a'\equiv _C a$ we have to check that $H\cap \overline {Cx} = G_{Cx}$ , this is similar to the argument above, using this time that $\pi (b)$ is $\mathbb {F}_p$ -independent over $\pi (\overline {C})$ . We have $a_i'-b_i\in G$ hence $\pi (a_i') = \pi (b_i)$ , for all $i\leq \lvert {x} \rvert $ .

Proof It is again a type to realise. Consider a tuple x from an elementary extension of the field K, such that x is of the same length as a and algebraically independent over $Cba$ . Let V be a $\mathbb {F}_p$ -vector space complement to $\overline {C}\oplus \langle {x} \rangle $ in $\overline {Cx}$ and set

$$ \begin{align*}H = G(\overline{Cab})+\langle {x-a} \rangle+V.\end{align*} $$

We check that the pair $(\overline {Cabx}, H)$ defines a consistent type over $Cab$ . We have to show that

$$ \begin{align*}H\cap \overline{Cab} = G(\overline{Cab}).\end{align*} $$

First, as $G(\overline {Cab})\subseteq \overline {{Cab}}$ , we have that

$$ \begin{align*}H\cap \overline{Cab} = G(\overline{Cab})+ (\langle {x-a} \rangle+V)\cap \overline{Cab},\end{align*} $$

so it suffices to show that $(\langle {x-a} \rangle +V)\cap \overline {Cab} = \{ {0} \}$ . Let $v\in V$ , if $\sum _i \lambda _i(x_i-a_i) +v\in \overline {Cab}$ then $\sum _i \lambda _i x_i + v\in \overline {Cab}$ . As $\overline {Cab}\cap \overline {Cx} = \overline C$ , $\sum _i \lambda _i x_i + v\in \overline {C}$ hence $v =0$ and, as x is $\mathbb {F}_p$ -independent over $\overline {C}$ , $\lambda _i = 0$ for all $i\leq \lvert {x} \rvert $ . The type is consistent by §2.1. We show now that

$$ \begin{align*}H\cap \overline{Cx} = G(\overline C )+V.\end{align*} $$

Again, as $V\subseteq \overline {{Cx}}$ ,

$$ \begin{align*}H\cap \overline{Cx} = V+\overline{Cx}\cap (G(\overline{Cab})+\langle {x-a} \rangle).\end{align*} $$

Let $g+ \sum _i \lambda _i (x_i-a_i) \in (G(\overline {Cab})+\langle {x-a} \rangle )\cap \overline {Cx}$ , then $g+\sum _i \lambda _i a_i \in \overline {Cab}\cap \overline {Cx} = \overline C$ and so applying $\pi $ gives $\sum _i \lambda _i \pi (a_i) \in \pi (\overline {C})$ hence $\lambda _i = 0$ for all $i\leq \lvert {x} \rvert $ . It follows that $\overline {Cx}\cap (G(\overline {Cab})+\langle {x-a} \rangle )=G(\overline {C})$ hence $H\cap \overline {Cx} = G(\overline C )+V$ . Assume that $a'$ realises this type, it is clear that $\pi (a)=\pi (a')$ and . By construction there exists $V'\subseteq \overline {Ca'}$ such that $\overline {Ca'} = \overline C \oplus \langle {a} \rangle \oplus V'$ and $G(\overline {Ca'}) = G(\overline {C})\oplus V'$ , so it follows that $\pi (\overline {Ca'}) = \pi (C)\oplus \langle {\pi (a')} \rangle $ .⊣

Proof We may assume that $\gamma $ is linearly independent over $\pi (\overline {C})$ and let $r_\gamma $ be a minimal representative of $\gamma $ over C. Let a and b be representatives of $\alpha $ and $\beta $ over $\overline {{Cr_\gamma }}$ . Using Lemma 3.2, there exists $a' \equiv _{Cr_\gamma } a$ such that $\pi (a') = \pi (b) = \beta $ . Let $\sigma $ be an automorphism of $(K,K/G)$ over $Cr_\gamma $ sending a on $a'$ . It is clear that $\sigma $ fixes $\gamma $ and sends $\alpha $ to $\beta $ hence $\alpha \equiv _{C\gamma } \beta $ .⊣

Proof For the first assertion, we may assume that $\gamma $ is an $\mathbb {F}_p$ -independent tuple over $\pi (\overline {C})$ . Let u be in $\mathrm {acl}(C\gamma )\cap K$ witnessed by an algebraic formula $\phi (x,c,\gamma )$ with $c\in C$ . Using twice Lemma 3.3, let $r_\gamma $ be a minimal representative of $\gamma $ over C, and $r_\gamma '$ a minimal representative of $\gamma $ over C such that . As u satisfies $\phi (x,c,\pi (r_\gamma ))$ and $\phi (x,c,\pi (r_\gamma '))$ , u belongs to $\overline {Cr_\gamma }\cap \overline {Cr_\gamma '} = \overline {C}$ (note that we don’t use the minimality here). The reverse inclusion being trivial, it follows that $\mathrm {acl}(C\gamma ) \cap K = \overline {C}$ .

For the second assertion, assume that $\alpha \notin \langle {\pi (\overline {C})\gamma } \rangle $ . By Corollary 3.4, any element in $K/G\setminus \langle {\pi (\overline {C})\gamma } \rangle $ has the same type as $\alpha $ over $C\gamma $ hence $\alpha \notin \mathrm {acl}(C\gamma )$ . The reverse inclusion being trivial, it follows that $\mathrm {acl}(C\gamma )\cap K/G = \langle {\pi (\overline {C})\gamma } \rangle $ .⊣

Proof (1) implies (2). Let $u_a\in \overline {a}$ and $u_b\in \overline {b}$ such that $u_a-u_b \in G$ . Then $\pi (u_a)=\pi (u_b) \in \pi (\overline {C})+\langle {\gamma } \rangle $ so there exists $u_c\in \overline {C}$ and $\lambda \in \mathbb {F}_p^{\lvert {\gamma } \rvert }$ such that for some (any) representatives $r^a$ and $r^b$ of $\gamma $ in $\overline {a}$ and $\overline {b}$ respectively, there exists $g_a\in G(\overline {a})$ , $g_b\in G(\overline {b})$ and an almost trivial sequence $\lambda \in \mathbb {F}_p^{\lvert {\gamma } \rvert }$ with

$$ \begin{align*} u_a = g_a + u_c +\sum_i \lambda_i r^a_i,&\\ u_b = g_b + u_c +\sum_i \lambda_i r^b_i.& \end{align*} $$

It follows that $u_a-u_b \in G(\overline {a})+G(\overline {b})+\langle {r^a - r^b} \rangle $ .

(2) implies (1). If $u_a\in \overline {a}$ and $u_b\in \overline {b}$ are such that $\pi (u_a) = \pi (u_b)$ , then $u_a-u_b \in G(\overline {a}+\overline {b})$ hence $u_a-u_b = g_a+g_b + \sum _i \lambda _i (r^a_i - r^b_i)$ (for an almost trivial sequence $\lambda \in \mathbb {F}_p^{\lvert {\gamma } \rvert }$ ). It follows that $u_a-g_a - \sum _i \lambda _i r^a_i\in \overline {a}\cap \overline {b} = \overline {c}$ , so $\pi (u_a) \in \pi (\overline {C}) +\langle {\gamma } \rangle $ .⊣

Proof Let x be an algebraically independent tuple over K of size $\lvert {d} \rvert $ , and define H on $K(x)$ to be $G(K) + \langle {x-d} \rangle $ . It is easy to see that $(\overline {K(x)}, H)$ defines a consistent type over K so let $r_\gamma $ be a realization of this type in an elementary extension $(K',G')$ of $(K,G)$ . We may assume that $(K',G')$ is $\kappa $ -saturated and $\kappa $ -homogeneous for some big enough $\kappa $ .

(1) Assume that $a\equiv _{C\gamma } b$ and let $\sigma $ be an automorphism of $(K',K'/G')$ over $C\gamma $ sending a on b. As before, we have that $G(\overline {ar_\gamma })= G(\overline {a})+\langle {r_\gamma - r^a} \rangle $ and $G(\overline {br_\gamma })= G(\overline {b})+\langle {r_\gamma - r^b} \rangle $ , for some (any) representatives $r^a, r^b$ of $\gamma $ in $\overline {a}$ , $\overline {b}$ respectively. Let $R_\gamma $ be an enumeration of $\overline {Cr_\gamma }$ and $R_\gamma ' = \sigma (R_\gamma )$ , $r_\gamma ' = \sigma (r_\gamma )$ . As , we have . Furthermore $G(\overline {ar_\gamma })= G(\overline {a})+\langle {r_\gamma - r^a} \rangle $ and $aR_\gamma \equiv _{C\gamma } bR_\gamma '$ , then $G(\overline {Cbr_\gamma '})= G(\overline {b})+\langle {r_\gamma ' - r^b} \rangle $ . By the claim, $\sigma ^{-1}\upharpoonright \overline {Cr_\gamma '}$ extends $\overline {Cr_\gamma 'b}$ with the identity on b hence $R_\gamma \equiv _{Cb\gamma } R_\gamma '$ . It follows that $aR_\gamma \equiv _{C\gamma } bR_\gamma $ . The other direction is trivial.

(2) From left to right. It is clear that . We want to show that $G(\overline {ar_\gamma }+\overline {br_\gamma }) = G(\overline {ar_\gamma })+G(\overline {br_\gamma })$ . Observe that $G(\overline {abr_\gamma }) = G(\overline {ab})+\langle {r^a-r_\gamma } \rangle $ for any tuple $r^a$ from $\overline {a}$ with $\pi (r^a)$ . Let $u\in \overline {ar_\gamma }$ and $v\in \overline {br_\gamma }$ . If $u+v\in G$ there exists $g_{ab}\in G(\overline {ab})$ and $\lambda \in \mathbb {F}_p^{\lvert {x} \rvert }$ such that $u+v = g_{ab} + \sum _i\lambda _i(r^a_i - r_{\gamma i})$ for an almost trivial tuple $\lambda $ . It follows that $g_{ab} \in (\overline {ar_\gamma } + \overline {br_\gamma })\cap \overline {ab} = \overline {a}+\overline {b}$ , using [Reference d’Elbée16, Lemma 5.16] with $T=\mathrm {ACF}$ . As and using Lemma 3.7, we have that $G(\overline {a}+\overline {b}) = G(\overline {a})+G(\overline {b})+\langle {r^a - r^b} \rangle $ . We deduce that $g_{ab} = g_a + g_b + \sum _i \mu _i(r^a_i - r^b_i)$ , for an almost trivial tuple $\mu $ . For all i, $r^a_i-r_{\gamma i}\in G(\overline {a r_\gamma })$ and $r_{\gamma i}-r^b_i \in G(\overline {br_\gamma })$ hence $g_{ab}= g_a + g_b + \sum _i \mu _i(r^a_i - r_{\gamma i}) + \sum _i \mu _i(r_{\gamma i} - r^b_i)\in G(\overline {ar_\gamma })+G(\overline {br_\gamma })$ . It follows that $u+v \in G(\overline {ar_\gamma })+G(\overline {br_\gamma })$ . The other inclusion being trivial we have $G(\overline {ar_\gamma }+\overline {br_\gamma }) = G(\overline {ar_\gamma })+G(\overline {br_\gamma })$ .

From right to left. First, hence by Transitivity and Monotonicity . By hypothesis, $G(\overline {ar_\gamma }+\overline {br_\gamma }) = G(\overline {ar_\gamma })+G(\overline {br_\gamma })$ . Furthermore $G(\overline {ar_\gamma }) = G(\overline a)+\langle {r_\gamma - r^a} \rangle $ and $G(\overline {br_\gamma }) = G(\overline b)+\langle {r_\gamma - r^b} \rangle $ . It is easy to see that

$$ \begin{align*}(G(\overline a)+ G(\overline b)+\langle {r_\gamma - r^a} \rangle+\langle {r_\gamma - r^b} \rangle)\cap (\overline{a}+\overline{b}) = G(\overline a)+ G(\overline b)+\langle {r^b - r^a} \rangle.\end{align*} $$

It follows that .

Proof Transitivity is just checking from the definition of . For Full Existence, assume the hypothesis and let $r_\gamma $ be a maximal representative as in Lemma 3.8. By Full Existence of in $(K,G)$ there exists $a'\equiv _{Cr_\gamma } a$ such that . Using again Lemma 3.8, $a'\equiv _{C\gamma } a$ and . For Independence theorem, we use the same strategy. Assume the hypothesis and let $r_\gamma $ be a maximal representative of $\gamma $ as in Lemma 3.8. From Lemma 3.8, we have that $c_1\equiv _{\overline {Cr_\gamma }} c_2$ and , , . As in $(K,G)$ satisfies -amalgamation over algebraically closed, sets there exists c such that $c\equiv _{\overline {Cr_\gamma }a} c_1$ , $c \equiv _{\overline {Cr_\gamma }b} c_2$ and . It follows that $c\equiv _{Ca\gamma } c_1$ , $c \equiv _{Cb\gamma } c_2$ , and by Lemma 3.8, .⊣

Proof If e is not in $\mathrm {dcl}^{eq}(E)$ , then there exists $e'\neq e$ such that $e'\equiv _E e$ . Let $\sigma $ be an automorphism of $\mathscr {M}^{eq}$ over E sending e on $e'$ . Let $b_1b_2 = \sigma (a_1a_2)$ . By Invariance, and $f(b_1)=f(b_2) = e'$ . By Extension there exists $b\equiv _{Eb_1}b_2$ such that . By Independent Consistency, there exists a such that $a\equiv _{Ea_1} a_2$ , $a \equiv _{Ea_2} b$ . From $a\equiv _{Ea_1} a_2$ follows that $f(a) = f(a_1) = e$ and from $a \equiv _{Ea_2} b$ follows that $f(a)\neq e$ , a contradiction.⊣

Fact 3.14. Let $\mathscr {M}$ be a saturated model, X a $0$ -definable set, $e\in \mathscr {M}$ , $E = \mathrm {acl}(e)\cap X$ and a tuple a from X. Then there is a tuple b from X such that

$$ \begin{align*}a\equiv_{Ee} b \mbox{ and } \mathrm{acl}(Ea)\cap \mathrm{acl}(Eb)\cap X = E.\end{align*} $$

Proof We work in $(K,G)^{eq}$ , seeing $(K,K/G)$ as a $0$ -definable subset. Suppose that e is an imaginary element, there is a tuple a from K and a $0$ -definable function f such that $e = f(a)$ . There exists $C = \overline {{C}}\subseteq K$ and $\gamma \subseteq K/G$ such that $C\langle {\pi (C)\gamma } \rangle = \mathrm {acl}^{eq}(e)\cap (K,K/G)$ . We may assume that $\gamma $ is $\mathbb {F}_p$ -linearly independent over $\pi (C)$ . As $\gamma \subseteq \mathrm {acl}^{eq}(e)\cap K/G \subseteq \mathrm {acl}^{eq}(a)\cap K/G$ we have that $\mathrm {acl}^{eq}(Ca\gamma )\cap (K,K/G) = \overline {Ca}\pi (\overline {Ca})$ and $\gamma \subseteq \pi (\overline {Ca})$ . By Fact 3.14 there exists $b\equiv _{C\gamma e} a$ such that

$$ \begin{align*}\mathrm{acl}^{eq}(Ca\gamma)\cap \mathrm{acl}^{eq}(Cb\gamma) \cap (K,K/G) = C\langle {\pi(C)\gamma} \rangle.\end{align*} $$

Again, $\mathrm {acl}^{eq}(Cb\gamma )\cap (K,K/G) = \overline {Cb}\pi (\overline {Cb})$ and $\gamma \subseteq \pi (\overline {Cb})$ . Furthermore $f(b) = e$ and

$$ \begin{align*}(\overline{Ca}\pi(\overline{Ca}))\cap (\overline{Cb}\pi(\overline{Cb})) = C\langle {\pi(C)\gamma} \rangle.\end{align*} $$

We construct a sequence $(a_i)_{i<\omega }$ such that

Start by $a_1 = a$ and $a_2 = b$ . Assume that $a_1,\dots ,a_n$ has already been constructed. We have that $a_{n-1} \equiv _{C\gamma } a_n$ so let $\sigma $ be a $C\gamma $ -automorphism of the monster such that $\sigma (a_{n-1}) = a_n.$ By Full Existence (Theorem 3.10) there exists $a_{n+1} \equiv _{Ca_n\gamma } \sigma (a_n)$ such that

It follows that

$$ \begin{align*}a_n a_{n+1} \equiv_{C\gamma} a_n\sigma(a_n)\equiv_{C\gamma} a_{n-1}a_n.\end{align*} $$

Let $(a_i)_{i<\omega }$ be such a sequence. In particular the following holds for all $i<j<k$

By Ramsey and compactness we may assume that $(a_i)_{i<\omega }$ is indiscernible over $C\gamma $ . As the three properties above hold for the whole sequence, it is in the Ehrenfeucht–Mostowski type of the sequence, and hence is still true for the indiscernible sequence. Note that $f(a_i) = e$ . We have that $(a_i)_{i<\omega }$ is totally indiscernible over C in the sense of $\mathrm {ACF}$ hence $a_1a_2a_3 \equiv ^{\mathrm {ACF}}_C a_1a_3a_2$ . Furthermore we have

, hence by Invariance

. By elimination of imaginaries in $\mathrm {ACF}$ it follows that

, since $\overline {{Ca}}\cap \overline {{Cb}} = \overline {{C}}$ . As $\pi (\overline {Ca_1})\cap \pi (\overline {Ca_2}) = \langle {\pi (C)\gamma } \rangle $ , we have that

As $f(a_1) = f(a_2) = e$ , we deduce from Lemma 3.12 that $e\in \mathrm {dcl}^{eq}(C\gamma )$ .⊣