Fact 2.4 ([Reference Chatzidakis4, Proposition 1.18] and [Reference Palacín and Wagner16, Theorem 3.6])

Let $\mathit{stp}(b/A)$ and $\mathit{stp}(b/C)$ be two $\mathbb {P}$ -analysable types.

1. (a) The type $\mathit{stp}\big (b/\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A)\cap \mathit{acl}^{\kern1.5pt \mathrm{eq}}(C)\big )$ is again $\mathbb {P}$ -analysable. In particular, so is $\mathit{stp}\big (b/\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A)\cap \mathit{acl}^{\kern1.5pt \mathrm{eq}}(b)\big )$ also $\mathbb {P}$ -analysable.

2. (b) Let $b_A$ be the maximal subset of $\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A,b)$ such that $\mathit{stp}(b_A /A)$ is almost $\mathbb {P}$ -internal. The tuple $b_A$ (in some fixed enumeration) dominates b over A, that is, for every set of parameters $D\supset A$ ,

   
   Furthermore, whenever $\mathit{acl}^{\kern1.5pt \mathrm{eq}}(D)\cap \mathit{acl}^{\kern1.5pt \mathrm{eq}}(A,b_A)=\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A)$ , so is
   $$ \begin{align*} \mathit{acl}^{\kern1.5pt \mathrm{eq}}(D)\cap\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A,b)=\mathit{acl}^{\kern1.5pt \mathrm{eq}}(A).\end{align*} $$

Proof We assume first that every stationary almost $\mathbb {P}$ -internal type is good, but the conclusion fails, witnessed by two tuples a and b. By Remark 2.2, we may assume

$$ \begin{align*} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset).\end{align*} $$

Thus, the type $\mathrm{stp}(b/a)$ is almost $\mathbb {P}$ -internal, but the type $\mathrm{stp}(b)$ is not. Note that $\mathrm{stp}(b)$ is $\mathbb {P}$ -analysable, by Fact 2.4.

Among all possible (imaginary) tuples in the ambient model take now $a'$ such that $\mathrm{stp}(b/a')$ is almost $\mathbb {P}$ -internal and

$$ \begin{align*}\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset)\end{align*} $$

with $\mathrm{U}(b_\emptyset /a')$ maximal. Since $\mathrm{stp}(b/a')$ is almost $\mathbb {P}$ -internal, there is a set of parameters A containing $a'$ with such that b is algebraic over $Ae$ , where e is a tuple of realizations of types (each one based over A) in $\mathbb {P}$ . Since each type in the family $\mathbb {P}$ is minimal, we may assume, after possibly enlarging A, that e and b are interalgebraic over A.

Let now $e'$ be a maximal subtuple of e independent from $b_\emptyset $ over A, so

Hence, the tuple b is algebraic over $Ae' b_\emptyset $ and

$$ \begin{align*}\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A,e')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b_\emptyset)\subset \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset).\end{align*} $$

Therefore $\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A,e')\cap \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b) =\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset )$ , by Fact 2.4.

Notice that $\mathrm{stp}(b/A, e')$ is almost $\mathbb {P}$ -internal, yet this does not yield any contradiction since $\mathrm{U}(b_\emptyset /A,e')=\mathrm{U}(b_\emptyset /a')$ . Choose now $b'$ realizing $\mathrm{stp}(b/A,e')$ independent from b over $A, e'$ . An easy forking computation yields

$$ \begin{align*} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset).\end{align*} $$

By the hypothesis we have that the almost $\mathbb {P}$ -internal type

$$ \begin{align*}\mathrm{stp}\big(b'/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A,e')\big)=\mathrm{stp}\big(b/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A,e')\big) \end{align*} $$

is good, so we deduce that $\mathrm{stp}(\mathrm{Cb}(b/A,e')/b')$ is almost $\mathbb {P}$ -internal. Remark that b is algebraic over $\mathrm{Cb}(b/A,e', b_\emptyset )$ and thus also algebraic over $b_\emptyset \mathrm{Cb}(b/A,e')$ .

Putting all of the above together, we conclude that the type $\mathrm{stp}(b/b')$ is almost $\mathbb {P}$ -internal. Since

$$ \begin{align*}\mathrm{U}(b_\emptyset/b')\geq \mathrm{U}(b_\emptyset/A, e',b') = \mathrm{U}(b_\emptyset/A,e')=\mathrm{U}(b_\emptyset/a'),\end{align*} $$

we deduce by the maximality of $\mathrm{U}(b_\emptyset /a')$ that $\mathrm{U}(b_\emptyset /b')= \mathrm{U}(b_\emptyset /A, e',b')$ , that is,

Hence , so , by Fact 2.4, contradicting that $\mathrm{stp}(b)$ is not almost $\mathbb {P}$ -internal.

For the other direction, we need to show that the almost $\mathbb {P}$ -internal type $\mathrm{stp}(a/b)$ is good, that is, that $\mathrm{stp}(\mathrm{Cb}(a/b)/a)$ is almost $\mathbb {P}$ -internal. We may assume that b equals the canonical base $\mathrm{Cb}(a/b)$ . Superstability yields that b is contained in the algebraic closure of finitely many b-conjugates of a. By transfer of internality to intersections, the type $\mathrm{tp}\big (a/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a)\cap \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)\big )$ is almost $\mathbb {P}$ -internal, so it follows that

$$ \begin{align*}\mathrm{tp}\big(b/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)\big)\end{align*} $$

is almost $\mathbb {P}$ -internal. Hence, the type $\mathrm{stp}(b/a)$ is almost $\mathbb {P}$ -internal, as desired.

Proof Assume that every stationary almost $\mathbb {P}$ -internal type is special. We want to show that

$$ \begin{align*}\mathrm{tp}\big(b/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_1)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_2)\big)\end{align*} $$

is almost $\mathbb {P}$ -internal, whenever both $\mathrm{stp}(b/a_1)$ and $\mathrm{stp}(b/a_2)$ are. By Remark 2.2, we may assume that

$$ \begin{align*} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_1)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_2)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset).\end{align*} $$

Note that the type $\mathrm{stp}(b)$ is $\mathbb {P}$ -analysable, by Fact 2.4, so recall that $b_\emptyset $ is the maximal almost $\mathbb {P}$ -internal subset of $\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)$ . As in the proof of Proposition 2.5 there is a set of parameters $A_1$ containing $a_1$ such that

and b is interalgebraic over $A_1$ with some tuple e of realizations of types (each one based over $A_1$ ) in $\mathbb {P}$ . Choosing a maximal subtuple $e'$ of e with

, it follows that b is algebraic over $b_\emptyset A_1 e'$ and that

$$ \begin{align*} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b_\emptyset)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A_1,e')\subset \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_1). \end{align*} $$

Hence

(⋆) $$ \begin{align} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A_1,e')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_2)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset), \end{align} $$

by Fact 2.4. Since the almost $\mathbb {P}$ -internal type $\mathrm{stp}(b / A_1 , e')$ is special, we have that

is almost $\mathbb {P}$ -internal. Therefore

is almost $\mathbb {P}$ -internal by Remark 2.2 and Proposition 2.5. Since

and b is algebraic over $b_\emptyset A_1 e'$ , the tuple b is algebraic over $\mathrm{Cb}(b / A_1 , e') b_\emptyset $ . In particular, the type

$$ \begin{align*} \mathrm{stp}\big(b/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(A_1, e')\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(a_2)\big) \end{align*} $$

is almost $\mathbb {P}$ -internal and hence so is $\mathrm{stp}(b)$ because of ( $\star $ ).

In order to prove the other direction, we want to show that the almost $\mathbb {P}$ -internal type $\mathrm{stp}(a/b)$ is special. Fix C some a set of parameters such that $\mathrm{stp}(a/C)$ is almost $\mathbb {P}$ -internal. By transfer of internality to quotients, the type

$$ \begin{align*}\mathrm{stp}\big(a/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(b)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(C)\big) \end{align*} $$

is almost $\mathbb {P}$ -internal and so is

since the canonical base $\mathrm{Cb}(a/b)$ is algebraic over finitely many b-conjugates of a.

Question 1. Are transfer of internality to intersections and to quotients equivalent properties for theories of finite rank?

Question 2. Does a theory of finite rank transfer internality to intersections, resp. to quotients, if every stationary real almost $\mathbb {P}$ -internal type is good, resp. special?

Proof We proceed by induction on the size n of the non-empty set A. For $n=1$ , there is nothing to prove. Assume A contains at least two elements, and choose a some element of A. Set $b=\sigma (a)$ . Inductively, we have that $\sigma $ maps the average $d_1$ of $A\backslash \{a\}$ to the average $d_2$ of $\sigma [A]\backslash \{b\}$ . Let $\varepsilon _1$ and $\varepsilon _2$ be the unique elements in P such that $\varepsilon _1\star d_1=a$ and $\varepsilon _2\star d_2=b$ . A straight-forward computation yields that $\frac {\varepsilon _1}{n}\star d_1$ , resp. $\frac {\varepsilon _2}{n}\star d_2$ , is the average of A, resp. of $\sigma [A]$ . Now the claim follows since $\sigma $ maps $\frac {\varepsilon _1}{n}$ to $\frac {\varepsilon _2}{n}$ .

Proof Since $\mathrm{RM}(a_i/\bar \alpha )=1$ , where $\bar {\alpha }=(\alpha _1,\ldots ,\alpha _n)$ , we may assume after possibly reordering that $\mathrm{RM}(a_1,\ldots ,a_m/\bar \alpha )=m=\mathrm{RM}(a_1,\ldots ,a_n/\bar \alpha )$ . In particular, the realization $a_i$ of the strongly minimal type $ \mathrm{tp}(a_i/\alpha _i)$ is algebraic over $\bar {\alpha },\hat {a}$ , where $\hat a=(a_1,\ldots ,a_m)$ . Set $b_i=(\alpha _i,b_{i}')$ the average of the finite set of $\{\bar {\alpha },\hat {a}\}$ -conjugates of $a_i$ . The element $b_i$ is definable over $\bar {\alpha },\hat {a}$ , by Lemma 3.6.

Therefore $a_i=(-b_{i}')\star b_i$ is definable over $\bar {\alpha },\hat {a}$ . Since the fibers of the projection $\pi $ are strongly minimal (see Remark 3.3), the type $\mathrm{tp}(\hat {a}/\bar {\alpha })$ is stationary, so we obtain the desired conclusion.

Proof Note that (a) is immediate since $\sigma $ is an automorphism of the full structure $(\mathbb C,\mathbb C \times \mathbb C)$ and every $\emptyset $ -definable set in $\mathcal M$ is $\emptyset $ -definable in $(\mathbb C,\mathbb C \times \mathbb C)$ . For (b), let a be an element in S algebraic over $B,P$ . The average b of the finite set of $B,P$ -conjugates of a is definable over $B,P$ , by Lemma 3.6. Since $\pi (b)=\pi (a)$ and the definable action $\star $ of P is transitive on each fiber, it follows that a is definable over $B,P$ , as desired. For (c), let $b_1$ and $b_2$ be two elements in the fiber of $\beta $ with distinct types over $A,P$ . Strong minimality of the fiber implies that either $b_1$ or $b_2$ is algebraic over $A,P$ . Now $b=\gamma _1\star b_1=\gamma _2\star b_2$ for some $\gamma _1$ and $\gamma _2$ in P, so b is algebraic over $A,P$ .

Proof Choose some generic element $\beta $ in P independent from $\alpha _1$ and set $a=\beta \star a_1=(\alpha _1,\beta )$ . Note that the Morley rank of a is two. If $a_1$ were not generic, then $a_1$ must be algebraic over the generic element $\alpha _1$ of P. Since $a=\beta \star a_1$ , it would follow that the generic element a of S is algebraic over P, which contradicts our assumption that the sort S is not almost P-internal. Hence $a_1$ is generic in S.

Now, we inductively assume that the tuple $\bar {a}_{<n}(a_1,\ldots ,a_{n-1})$ consists of generic independent elements and want to show that . Assume otherwise that . Note that $\alpha _n$ is not algebraic over $\bar {a}_{<n}$ , by Lemma 3.8(a), since $\alpha _n$ is not algebraic over $\bar {\alpha }_{<n}$ . Thus , so $a_n$ is algebraic over $\alpha _n \bar {a}_{<n}$ . Choose now some element $\gamma $ in P generic over $(\alpha _1,\ldots , \alpha _n)$ and set $c=(\alpha _n,\gamma )=\gamma \star a_n$ in S. Note that c is algebraic over $\bar {a}_{<n}P$ . Observe that $\mathrm{RM}(c/\bar {a}_{<n})=2$ , by the choice of $\gamma $ , so we conclude that S is almost P-internal, which gives the desired contradiction.

Proof Clearly, it suffices to show that, whenever the real type $\mathrm{stp}(a/B)$ is almost P-internal, with a a single element in S, then $\alpha =\pi (a)$ is algebraic over B. Choose a set of parameters $B_1$ with such that a is algebraic over $B_1,P$ . We need only show that $\alpha $ is algebraic over $B_1$ . Otherwise, choose an element $a_1$ of S in the fiber of $\alpha $ generic over $B_1$ . The elements a and $a_1$ are interdefinable over P, so $a_1$ is algebraic over $B_1, P$ , contradicting that S is not almost P-internal.

Proof The equivalence in the statement holds trivially, since $G\star a$ is a coset of the P-definable group $G\star 0_{S^n}$ .

The type q is clearly P-internal and fundamental, because $\pi (a)=(\alpha _1,\ldots ,\alpha _n)$ is definable over $\ulcorner G\star a \urcorner $ . Given a realization b of q, it lies in $G\star a$ , so $b=g\star a$ , for some g in G. With the above identification, we deduce that the binding group of q is a subgroup of G. In case $a_1,\ldots ,a_n$ are generic independent, we have equality of these two groups: the element $a_k$ , for $1\leq k\leq n$ , is not algebraic over $\bar {a}_{<k},P$ , since S is not almost P-internal, so Lemma 3.8(c) yields that all elements in the fiber $\pi ^{-1}(\alpha _k)$ have the same type over $\bar {a}_{<k},P$ . For any $g=(g_1,\ldots , g_n)$ in G we can recursively construct an automorphism $\sigma $ in $\mathrm{Aut}(\mathcal {M}/P)$ with $\sigma (a_k)=g_k \star a_k$ for $1\leq k\leq n$ . By the previous equivalence, the automorphism $\sigma $ permutes $G\star a$ and therefore induces an automorphism in the binding group of q.

That every binding group is of the claimed form follows from the discussion before the proposition. Let us now conclude by showing that every definable subgroup G of $(\mathbb {C}^n,+)$ appears as a binding group, if S is not almost P-internal. Choose a generic independent tuple a in $S^n$ . By the above, the binding group of the P-internal type $\mathrm{stp}(a/\ulcorner G\star a\urcorner )$ equals G.

Proof Set $\alpha =\pi (a)$ and $\beta =\pi (b)$ , for two generic independent elements a and b in the sort S. Fix a derivation D with kernel $\mathbb {Q}^{\text {alg}}$ and let G be the definable subgroup of $(P^2,+)$ given by $D(\beta )x-D(\alpha )y=0$ . Let us assume for a contradiction that the definable set $G\star (a,b)$ has a real canonical parameter e. By hypothesis, the induced map $\sigma _{D}$ is in $\mathrm{Aut}(\mathcal {M}/P)$ . By Proposition 3.11, the automorphism $\sigma _D$ permutes the set $G\star (a,b)$ , since $\sigma _D(a,b)=(D(\alpha )\star a ,D(\beta )\star b)$ lies in $G\star (a,b)$ because the tuple $(D(\alpha ),D(\beta ))$ is an element of G. By the choice of the derivation D, every real element fixed by $\sigma _D$ belongs to the set $P\cup \pi ^{-1}(\mathbb {Q^{\text {alg}}})$ . In particular, we deduce that the tuple e lies in this set. Thus, the definable set $G\star (a,b)$ is permuted by every automorphism induced by a derivation, because every derivation vanishes on $\mathbb {Q^{\text {alg}}}$ (note that we no longer need that the kernel of the derivation is exactly $\mathbb {Q}^{\text {alg}}$ ). Choose now a derivation $D_1$ with $D_1(\alpha )=1$ and $D_1(\beta )=0$ , and note that $\sigma _{D_1}$ does not permute $G\star (a,b)$ , since $(1,0)$ is not an element of G, which gives the desired contradiction.

Proof We will mimic the proof of Lemma 4.1. Assume for a contradiction that the sort S is not P-internal and choose two generic independent elements a and b in S. Consider now the definable set $G\star (a,b)$ where G is given by the equation $F_{\sigma }(\beta )x-F_{\sigma }(\alpha )y=0$ , for some suitable automorphism $\sigma $ to be determined later on. If we can construct the automorphism $\sigma $ (which plays the role of $\sigma _D$ in the proof of Lemma 4.1) in $\mathrm{Aut}(\mathcal {M}/P)$ such that the only real elements fixed by $\sigma $ are those definable over P, we conclude as before that the real canonical parameter e of the definable set $G\star (a,b)$ is definable over P. This yields immediately the desired contradiction, if S were not P-internal, for there is an automorphism $\tau $ in $\mathrm{Aut}(\mathcal {M}/P)$ which fixes b and moves a, but $\tau (a,b)=(F_\tau (\alpha )\star a, b)$ does not belong to $G\star (a,b)$ , since $(0,0)\neq (F_\tau (\alpha ),0)$ is not in G, since $F_{\sigma }(\beta )\cdot F_{\tau }(\alpha )\neq 0$ .

Hence, we need only show in the rest of the proof that there exists such an automorphism $\sigma $ .

Choose a transcendence basis $\bar {\alpha }=(\alpha _i)_{i<2^{\aleph _0}}$ of the algebraically closed field $\mathbb {C}$ and set

$$ \begin{align*}\bar{a}=(a_i)_{i<2^{\aleph_0}}=\big((\alpha_i,0)\big)_{i<2^{\aleph_0}}\end{align*} $$

for $i<2^{\aleph _0}$ . We construct a (possibly empty) subtuple

$$ \begin{align*}\bar{b}=(b_i)_{i<\kappa}=\big((\beta_i,0)\big)_{i<\kappa}\end{align*} $$

of $P\times \{0\} =\big ((\varepsilon ,0)\big )_{\varepsilon \in \mathbb C}$ in the following way: Assume $(b_i)_{i<\lambda }$ has already been constructed. If some element $(\varepsilon ,0)$ is not algebraic over $\bar a, (b_i)_{i<\lambda }$ , then choose it least possible according to some fixed enumeration of $\mathbb C$ and set $b_{\lambda }=(\varepsilon ,0)$ . Otherwise set $\kappa =\lambda $ .

Set $\bar {\beta }=(\beta _i)_{i<\kappa }$ . For $i<\kappa $ , we denote by ${\langle \alpha \rangle }_i$ the unique subtuple of $\bar {\alpha }$ of smallest length such that $\beta _i$ is algebraic over ${\langle \alpha \rangle }_i$ . Write $\mathcal X$ for the set of all finite subtuples of $\bar {\alpha }$ and consider the maps

$$ \begin{align*} \begin{array}{rccc} \Phi: & \mathcal X & \rightarrow & 2^{\aleph_0} \\[2mm] & (\alpha_{i_1},\ldots,\alpha_{i_n})&\mapsto& \max(i_1,\ldots,i_n) \end{array}\end{align*} $$

and

$$ \begin{align*}F:\{ \alpha_i,\beta_j \ |\ i<2^{\aleph_0}, j<\kappa \}\rightarrow \{ \alpha_i \ |\ i<2^{\aleph_0}\}\end{align*} $$

defined by

$$ \begin{align*} F(\alpha_i)=\alpha_{\omega^{i+1}} \ \ \text{ and } \ \ F(\beta_j)=\alpha_{\omega^{\max\left(j,\Phi({\langle \alpha \rangle}_j)\right)+1}+\omega^{j}}. \end{align*} $$

It follows inductively from Lemma 3.8(c) that

$$ \begin{align*}\bar{a},\bar{b} \equiv_{P} F(\bar{\alpha})\star\bar{a},F(\bar{\beta})\star\bar{b},\end{align*} $$

so there is an automorphism $\sigma $ in $\mathrm{Aut}(\mathcal {M}/P)$ with

$$ \begin{align*} \sigma(a_i)=F(\alpha_i)\star a_i \ \ \text{ and } \ \ \sigma(b_j)=F(\beta_j)\star b_j. \end{align*} $$

Proof of the Claim Assume that the element $(\gamma ,c')$ of S is fixed by $\sigma $ . Note that the automorphism $\sigma $ also fixes $c=(\gamma ,0)$ , for $(\gamma ,c')=c'\star c$ . Thus, it suffices to show that c is definable over P. By construction of the tuples $\bar a$ and $\bar b$ , the element c must be algebraic over $\bar a,\bar b$ and therefore definable over $\bar a, \bar b, P$ , by Lemma 3.8(b). Choose subtuples of least possible length

$$ \begin{align*} \hat{a}=(a_{i_1},\ldots,a_{i_m}) \text{ of } \bar a \ \ \text{ and } \ \ \hat{b}=(b_{j_1},\ldots,b_{j_n}) \text{ of } \bar b \end{align*} $$

such that c is definable over $\hat {a},\hat {b},P$ . We need to show that $m=n=0$ . In order to reach a contradiction, assume that $\max (n,m)>0$ . Note that every element in the fiber of $\gamma $ is definable over $\hat {a},\hat {b},P$ . The type

$$ \begin{align*} \mathrm{tp}(\hat{a},\hat{b},c/\hat{\alpha},\hat{\beta},\gamma) \end{align*} $$

is fundamental and stationary by Lemma 3.7. Its binding group G is a definable additive subgroup of $\mathbb {C}^{m+n+1}$ , by Proposition 3.11. Note that $\gamma $ must be algebraic over $\hat {\alpha },\hat {\beta }$ : otherwise, Lemma 3.9 yields that , so $\mathrm{stp}(c)$ is P-internal (for c is definable over $\hat {a},\hat {b},P$ ) and hence so is S, contradicting our assumption.

Hence, the binding group G of $p=\mathrm{tp}(\hat {a},\hat {b},c/\hat {\alpha },\hat {\beta },\gamma )$ is definable over

$$ \begin{align*}A=\mathrm{acl}(\hat{\alpha}, {\langle \alpha \rangle}_{j_1},\ldots,{\langle \alpha \rangle}_{j_n})\supset \{ \hat{\alpha},\hat{\beta}\}.\end{align*} $$

By the discussion before Proposition 3.11, the group G is given by a system $\mathcal {G}$ of linear equations of the form

$$ \begin{align*} \lambda_{1} \cdot x_{1}+\dots+\lambda_{m+n+1} \cdot x_{m+n+1}=0, \end{align*} $$

with coefficients $\lambda _{i}$ in A and the tuple

$$ \begin{align*} ( F(\alpha_{i_1}),\ldots, F(\alpha_{i_m}), F(\beta_{j_1}),\ldots, F(\beta_{j_n}), 0) \end{align*} $$

is a solution, since

$$ \begin{align*} \sigma(a_i)=F(\alpha_i)\star a_i, \ \ \sigma(b_j)=F(\beta_j)\star b_j \ \ \text{ and } \ \ \sigma(c)=0\star c. \end{align*} $$

Set now $\varepsilon = \Phi \big ( \hat {\alpha }{^{\frown }} {\langle \alpha \rangle }_{j_1}{{}^\frown } \cdots {}^\frown {\langle \alpha \rangle }_{j_n}\big )$ . If $\alpha _\varepsilon =\alpha _{i_k}$ for some $1\le k\le m$ , denote $i(\varepsilon )=i_k=\varepsilon $ . Otherwise set $i(\varepsilon )=j_\ell $ if $1\le \ell \le n$ is the least index such that $\alpha _\varepsilon $ is an element in the tuple ${\langle \alpha \rangle }_{j_\ell }$ .

Observe that there is a linear equation in the system $\mathcal {G}$ such that the coefficient $\lambda _{i(\varepsilon )}$ is non-trivial, for otherwise every automorphism in $\mathrm{Aut}(\mathcal {M}/P)$ fixing all coordinates except (possibly) the element $d_{i(\varepsilon )}$ , which is the $i(\varepsilon )^{\text {th}}$ -coordinate of the tuple $(\hat {a},\hat {b})$ , must also fix c, contradicting the minimality of m and n. The set

$$ \begin{align*} B=\{ F(\alpha_{i_1}),\ldots, F(\alpha_{i_m}), F(\beta_{j_1}),\ldots, F(\beta_{j_n}) \} \end{align*} $$

consists of distinct elements, since F clearly is injective. Therefore, if suffices to show that the element $F(d_{i(\varepsilon )})$ is not algebraic over

$$ \begin{align*} A\cup \big(B\backslash \{ F(d_{i(\varepsilon)}) \}\big) \end{align*} $$

to reach the desired contradiction. For this we need only show that the element $F(d_{i(\varepsilon )})$ is not contained in the set

$$ \begin{align*} \{ \hat{\alpha}, {\langle \alpha \rangle}_{j_1},\ldots,{\langle \alpha \rangle}_{j_n} \}. \end{align*} $$

If $d_{i(\varepsilon )}=\alpha _{i(\varepsilon )}$ , we obtain the result since

$$ \begin{align*} \Phi(F(d_{i(\varepsilon)}))&=\Phi(F(\alpha_\varepsilon)) \\ &=\Phi(\alpha_{\omega^{\varepsilon+1}}) \\ &=\omega^{\varepsilon+1}\geq\varepsilon+1 \\ &>\varepsilon = \Phi \big ( \hat {\alpha }{^{\frown }} {\langle \alpha \rangle }_{j_1}{{}^\frown } \cdots {}^\frown {\langle \alpha \rangle }_{j_n}\big ). \end{align*} $$

Otherwise $d_{i(\varepsilon )}=\beta _{i(\varepsilon )}$ , so

$$ \begin{align*} \Phi(F(d_{i(\varepsilon)}))&=\Phi(F(\beta_{i(\varepsilon)}))\\ &= \omega^{\max\left({i(\varepsilon)},\Phi({\langle \alpha \rangle}_{i(\varepsilon)})\right)+1}+\omega^{{i(\varepsilon)}} \\ &> \omega^{ \Phi({\langle \alpha \rangle}_{i(\varepsilon)}) +1 } = \omega^{\varepsilon+1}, \end{align*} $$

and we conclude the result analogous to the first case.

Proof Assume that the tuple e is a real canonical parameter of the set $\{a,b\}$ , with $a=(\alpha ,0)$ and $b=(\beta ,0)$ . We will first show that every element c in S appearing in the real tuple e is definable over $a\oplus b,P$ . Since the canonical embedding

$$ \begin{align*}\mathrm{Aut}(\mathcal{M}_0/P) \hookrightarrow \{F:\mathbb{C}\rightarrow\mathbb{C} \text{ additive}\}\end{align*} $$

is an isomorphism (see Example 3.2), every additive map vanishing on $\alpha $ and $\beta $ must vanish on $\pi (c)$ . We deduce that the projection $\pi (c)$ is a linear combination of $\alpha $ and $\beta $ with rational coefficients, so $\pi (c)=\lambda \cdot \alpha +\mu \cdot \beta $ for some rational numbers $\lambda $ and $\mu $ . Note that there is a field automorphism $\tau $ sending $\alpha $ to $\beta $ and $\beta $ to $\alpha $ , so the automorphism $\sigma =(\tau , \tau \times \tau )$ of Lemma 3.8(a) permutes a and b, hence it fixes c. Therefore,

$$ \begin{align*} \lambda\cdot\alpha+\mu\cdot\beta=\pi(c)=\sigma(\pi(c))= \sigma(\lambda\cdot\alpha+\mu\cdot\beta) =\lambda\cdot\beta+\mu\cdot\alpha, \end{align*} $$

so $\lambda =\mu $ , since $\alpha $ and $\beta $ are generic independent. We deduce that c is definable over $a\oplus b,P$ , because every additive map vanishing on the element $\alpha +\beta $ vanishes on $\lambda \cdot (\alpha +\beta )=\pi (c)$ .

Choose now an additive map F with $F(\alpha )=1$ and $F(\beta )= {-}1$ . The induced automorphism $\sigma _F$ in $\mathrm{Aut}(\mathcal {M}/P)$ with $\sigma _F (x)=F(\pi (x))\star x$ for elements x in S fixes $a\oplus b$ , because

$$ \begin{align*} \sigma_F(a\oplus b)=\sigma_F(a)\oplus \sigma_F(b)=(1\star a)\oplus(-1\star b)=(\alpha,1)\oplus(\beta,-1)=a\oplus b. \end{align*} $$

Therefore $\sigma _F$ fixes the real tuple e, which gives the desired contradiction, since $\sigma _F$ does not permute the set $\{(\alpha , 0),(\beta , 0)\}$ .

Proof Let A be the finite set $\{\bar {a}_1,\ldots ,\bar {a}_n\}$ of real m-tuples. Every function $\Phi :\{1,\ldots , m\}\longrightarrow \{P,S\}$ determines a subset $A_{\Phi }$ of A, according to whether the $j^{\text {th}}$ -coordinate lies in P or S. Every automorphism permuting A permutes each $A_{\Phi }$ , so we may assume that for every tuple in A, the coordinates have the same configuration (given by the function $\Phi $ ).

Since the canonical parameter is only determined up to interdefinability, we may further assume (after possibly permuting the coordinates) that there is a natural number $0\leq k\leq m$ such that for each tuple $\bar {a}_i$ in A:

• • The $j^{\text {th}}$ -coordinate $a_{i}^{j}$ lies in S for $1\leq j\leq k$ .

• • The $\ell ^{\text {th}}$ -coordinate $a_{i}^{\ell }$ lies in P for $k< \ell \leq m$ .

For every coordinate $1\leq j\leq k$ set $A^j=\{a_{i}^{j} \ | \ 1\leq i\leq n \}$ and $d_{i}^{j}$ the average of the subset $A^j \cap \pi ^{-1}(\pi (a_{i}^{j}))$ . For $1\leq i\leq n$ let now $\varepsilon _{i}^{j}$ be the unique element in P with $a_{i}^{j}=\varepsilon _{i}^{j} \star d_{i}^{j}$ . Consider the tuples $\varepsilon _{i}=(\varepsilon _{i}^{1},\ldots ,\varepsilon _{i}^{k})$ and

$$ \begin{align*} \alpha_i=(\pi(a_{i}^{1}),\ldots,\pi(a_{i}^{k}),a_{i}^{k+1},\ldots,a_{i}^{m})\end{align*} $$

in P. We need only show that the tuple

$$ \begin{align*} e=\big( \ulcorner \{(\varepsilon_{1},\alpha_{1}),\ldots, (\varepsilon_{n},\alpha_{n}) \}\urcorner,\ulcorner \{d_{1}^{1},\ldots,d_{n}^{1}\}\urcorner,\ldots,\ulcorner \{d_{1}^{k},\ldots,d_{n}^{k}\}\urcorner \big) \end{align*} $$

is a canonical parameter of A. Note that e is a real tuple since the sort P with the full field structure eliminates imaginaries and all the sets $\{d_{1}^{j},\ldots ,d_{n}^{j}\}$ have real canonical parameters, by our assumption.

Let $\sigma $ be an automorphism. If $\sigma $ permutes the set A, Lemma 3.6 yields that $\sigma $ permutes each set $\{d_{1}^{j},\ldots ,d_{n}^{j}\}$ since the image of $A^j \cap \pi ^{-1}(\pi (a_{i}^{j}))$ under $\sigma $ is $A^j \cap \pi ^{-1}(\pi (a_{i(\sigma)}^{j}))$ for some index $i(\sigma )$ with $\sigma (a_{i}^{j})=a_{i(\sigma )}^{j}$ and $\sigma (\alpha _i)=\sigma (\alpha _{i(\sigma )})$ . Therefore $\sigma (\varepsilon _{i})=\varepsilon _{i(\sigma )}$ , since $\sigma (d_{i}^{j})=d_{i(\sigma )}^{j}$ . Hence $\sigma $ fixes e.

Assume now that $\sigma $ fixes the tuple e. The tuple $\alpha _i$ is mapped to $\alpha _{i(\sigma )}$ and

$$ \begin{align*} \sigma(a_{i}^{j})=\sigma(\varepsilon_{i}^{j}) \star \sigma(d_{i}^{j})=\varepsilon_{i(\sigma)}^{j}\star \sigma(d_{i}^{j}). \end{align*} $$

It suffices to show that $\sigma (d_{i}^{j})=d_{i(\sigma )}^{j}$ to conclude that $\sigma $ permutes A. This follows immediately from

$$ \begin{align*}\pi(\sigma(d_{i}^{j}))=\sigma(\pi(d_{i}^{j}))= \sigma(\alpha_{i}^{j})=\alpha_{i(\sigma)}^{j},\end{align*} $$

since $\sigma $ permutes the set $\{d_{1}^{j},\ldots ,d_{n}^{j}\}$ .

Proof By Proposition 4.4, we need only show that every subset $\{a_1,\ldots ,a_n\}$ of S, with pairwise distinct projections $\pi (a_i)=\alpha _i$ , has a real canonical parameter.

For $1\leq i\leq n$ lift the $i^{\text{th}}$ -symmetric function to S:

(♠) $$ \begin{align} b_i=\sum_{1\leq j_1<\dots<j_i\leq n}a_{j_1}\otimes\cdots\otimes a_{j_i}.\end{align} $$

We claim that the tuple $b=(b_1,\ldots ,b_n)$ is a canonical parameter of the set $A=\{a_1,\ldots ,a_n\}$ . If the automorphism $\sigma $ permutes A, then it clearly fixes b. Assume now that $\sigma $ fixes the tuple b. Since

$$ \begin{align*} \beta_i=\sum_{1\leq j_1<\dots<j_i\leq n}\alpha_{j_1}\cdots\alpha_{j_i},\end{align*} $$

the tuple $(\beta _1,\ldots ,\beta _n)$ encodes the finite set $\{\alpha _1,\ldots ,\alpha _n\}$ . In particular the automorphism $\sigma $ induces a permutation of the set $\{1,\ldots ,n\}$ (which we will also denote by $\sigma $ ) such that

$$ \begin{align*} \sigma(a_i)=\lambda_{\sigma(i)}\star a_{\sigma(i)} \end{align*} $$

for some $\lambda _{\sigma (i)}$ in P. We need only show that each $\lambda _{j}$ equals $0$ .

Write each element $a_i$ of A as $a_i=(\alpha _i,a_{i}')$ , and similarly $b_i=(\beta _i,b_{i}')$ . In the full structure $\mathcal {\widetilde M}$ the definable condition ( $\spadesuit $ ) is equivalent to

$$ \begin{align*} \beta_i=\sum_{1\leq j_1<\dots<j_i\leq n}\alpha_{j_1}\cdots\alpha_{j_i}\end{align*} $$

together with the system of linear equations:

$$ \begin{align*} \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \sum\limits_{j\neq 1}\alpha_j & \sum\limits_{j\neq 2}\alpha_j & \cdots & \sum\limits_{j\neq n}\alpha_j \\ \sum\limits_{\substack{j_1<j_2 \\ j_1,j_2\neq 1}}\alpha_{j_1} \alpha_{j_2} & \sum\limits_{\substack{j_1<j_2 \\ j_1,j_2\neq 2}}\alpha_{j_1} \alpha_{j_2} & \cdots & \sum\limits_{\substack{j_1<j_2 \\ j_1,j_2\neq n}}\alpha_{j_1} \alpha_{j_2} \\ \vdots & \vdots & \ddots & \vdots \\ \prod\limits_{j\neq 1}\alpha_j & \prod\limits_{j\neq 2}\alpha_j & \cdots & \prod\limits_{j\neq n}\alpha_j \end{pmatrix} \begin{pmatrix} a_1' \\ a_2' \\ \vdots \\ \vdots \\ \vdots \\ a_n' \end{pmatrix} = \begin{pmatrix} b_1' \\ b_2' \\ \vdots \\ \vdots \\ \vdots \\ b_n' \end{pmatrix} \end{align*} $$

Since

$$ \begin{align*} b_i = \sigma(b_i)=\sum_{1\leq j_1<\dots<j_i\leq n}(\lambda_{\sigma^{-1}(j_1)}\star a_{j_1})\otimes\cdots\otimes (\lambda_{\sigma^{-1}(j_i)}\star a_{j_i}), \end{align*} $$

we conclude that $\lambda _1 = \cdots = \lambda _n=0$ , because the above matrix has determinant $\prod _{i<j}(\alpha _i-\alpha _j)\neq 0$ .

Fact 5.1. Let G be a definable group in a theory with the CBP. If the type $p=stp(g/B)$ of an element g of G has finite stabilizer, then p is almost internal to the family of all non-locally modular minimal types.

Proof Choose two generic independent elements $\alpha $ and $\beta $ in P and consider the elements $a=(\alpha ,0),\ b=(\beta ,0)$ and $c=(\alpha \cdot \beta ,0)$ in S. We first show that c is definable over $a,b$ . So, let $\sigma $ be an automorphism of $\mathcal {M}$ fixing a and b. The restriction $\sigma _{|P}$ is a field automorphism and induces an automorphism $\widetilde \sigma =(\sigma _{|P},\sigma _{|P}\times \sigma _{|P})$ of $\mathcal M$ , by Lemma 3.8(a). Note that the automorphism ${\widetilde \sigma }^{-1}\circ \sigma $ fixes $P,a,b$ . By our assumption, it corresponds to a derivation D with $D(\alpha )=0=D(\beta )$ . In particular

$$ \begin{align*}{\widetilde\sigma}^{-1}\circ \sigma(c)=D(\alpha\cdot\beta)\star c=c,\end{align*} $$

so

$$ \begin{align*} \sigma(c) =\widetilde\sigma\big(({\widetilde\sigma}^{-1}\circ \sigma)(c)\big) =\widetilde{\sigma}(c)=c, \end{align*} $$

as desired.

By Fact 5.1, we need only show that $\mathrm{stp}(a,b,c)$ has trivial stabilizer (since S is not almost P-internal). As in the discussion above, it follows that every tuple $(g_1,g_2,g_3)$ in the stabilizer must have trivial projection onto the first coordinates, so write $g_i=\gamma _i\star 0_S$ . Note that $g_i$ and $\gamma _i$ are interdefinable, so we may assume that

We have that $(\gamma _1\star a, \gamma _2\star b) \equiv _P (a,b)$ by Lemma 3.8(c). Hence, there is an automorphism $\sigma $ fixing P pointwise (and hence a derivation $D_\sigma $ , by assumption) such that $D_\sigma (\alpha ) =\gamma _1$ and $D_\sigma (\beta ) =\gamma _{2}$ . As

$$ \begin{align*} (a\oplus g_1, b\oplus g_2, c\oplus g_3)=(\gamma_1\star a, \gamma_2\star b, \gamma_3\star c) \equiv (a,b, c),\end{align*} $$

and both $\sigma (c)$ and $\gamma _3\star c$ are definable over $\gamma _1\star a,\gamma _2\star b$ by the same formula, we deduce that

$$ \begin{align*} \gamma_3=D_\sigma(\alpha\cdot\beta)= \gamma_1\cdot\beta+ \alpha\cdot\gamma_2.\end{align*} $$

The independence

yields that all $\gamma _i$ ’s are trivial, as desired.

Proof Recall that we need only consider real types over models in order to deduce that the CBP holds. Let $p(x)$ be the type of some finite real tuple $\bar {a}$ of length k over an elementary substructure N. In order to show that the type $\mathrm{stp}(\mathrm{Cb}(p)/\bar {a})$ is almost P-internal, choose a formula $\varphi (x;\bar {b},\gamma )$ in p of least Morley rank and Morley degree one, where $\bar {b}$ is a tuple of elements in $S\cap N$ and $\gamma $ is a tuple of elements in $P \cap N$ .

We claim that every automorphism in $\mathrm{Aut}(\mathcal {M}/P,\bar {a})$ fixes the canonical base $\mathrm{Cb}(p)$ , which is (interdefinable with) the canonical parameter $\ulcorner \text {d}_{p} x \varphi (x;y)\urcorner $ . For this, it suffices to show that every such automorphism $\sigma $ sends the tuple $\bar {b}$ to another realization of the formula $\text {d}_{p} x \varphi (x; y_1, \gamma )$ .

Write $\bar {a}=(a_1,\ldots ,a_k)$ and

$$ \begin{align*} \alpha_i=\begin{cases} \pi(a_i), \text{ if}\ a_i\ \text{is in}\ S,\\ a_i \text{ otherwise.} \end{cases} \end{align*} $$

For $\bar {b}=(b_1,\ldots ,b_n)$ , set $\beta _i=\pi (b_i)$ . We may assume (after possibly reordering) that $(\beta _1,\ldots ,\beta _m)$ is a maximal subtuple of $\bar {\beta }$ which is $\mathbb {Q}$ -linearly independent over $\bar {\alpha }$ . So,

$$ \begin{align*} \beta_j = \sum\limits_{i=1}^{m}q_i\cdot\beta_i + \sum\limits_{i=1}^{k}r_i\cdot\alpha_i \end{align*} $$

for $m+1\leq j\leq n$ and rational numbers $q_i$ and $r_i$ . In order to show that $\bar {b}$ is mapped by the automorphism $\sigma $ of $\mathrm{Aut}(\mathcal {M}/P,\bar {a})$ to another realization of the formula $\text {d}_{p} x \varphi (x;y_1, \gamma )$ , it suffices to show that

$$ \begin{align*} N\models \forall \varepsilon_1,\ldots,\varepsilon_m \in P\ \ \text{d}_{p} x \varphi(x;\bar{\varepsilon}\star\bar{b},\gamma), \end{align*} $$

where $\bar {\varepsilon }=(\varepsilon _1,\ldots ,\varepsilon _n)$ with

$$ \begin{align*} \varepsilon_j = \sum\limits_{i=1}^{m}q_i\cdot\varepsilon_i \end{align*} $$

for $m+1\leq j\leq n$ . Indeed: since N is an elementary substructure of $\mathcal {M}$ , the above implies that

$$ \begin{align*} \mathcal{M}\models \forall \varepsilon_1,\ldots,\varepsilon_m \in P\ \ \text{d}_{p} x \varphi(x,\bar{\varepsilon}\star\bar{b},\gamma), \end{align*} $$

so we deduce from

$$ \begin{align*} F_\sigma (\beta_j) = \sum\limits_{i=1}^{m}q_i\cdot F_\sigma (\beta_i) + \sum\limits_{i=1}^{k}r_i\cdot F_\sigma (\alpha_i) = \sum\limits_{i=1}^{m}q_i\cdot F_\sigma (\beta_i) +0\end{align*} $$

for $m+1\leq j\leq n$ , that $\sigma (\bar b)= F_\sigma (\bar b)\star \bar b$ realizes $\text {d}_{p} x \varphi (x;\bar y_1,\gamma )$ , as desired.

So, let $\varepsilon _1,\ldots ,\varepsilon _m$ be in $P\cap N$ and set $\varepsilon _j = \sum _{i=1}^{m}q_i\cdot \varepsilon _i $ for $m+1\leq j\leq n$ . Choose an additive map G vanishing on $\alpha _i$ for $1\leq i\leq k$ and with $G(\beta _i)=\varepsilon _i$ for $1\leq i \leq m$ . Hence

$$ \begin{align*}G(\beta_j)= \sum\limits_{i=1}^{m}q_i\cdot\varepsilon_i , \end{align*} $$

so the image of $\bar b$ under the automorphism $\sigma _G$ induced by G lies in N. Hence $\sigma _{G}(\bar {b})=\bar \varepsilon \star \bar b$ realizes $\text {d}_{p} x \varphi (x,y_1,\gamma )$ since $\sigma _{G}(\bar {a})=\bar {a}$ , as desired.

Proof Note that $\mathcal {Q}$ -internality coincides with P-internality. Moreover, the binding group relative to $\mathcal {Q}$ is a subgroup of the binding group relative to P, which by Proposition 3.11 is a definable subgroup of some $(\mathbb {C}^n,+)$ . So it suffices to show that every definable subgroup G of $(\mathbb {C}^n,+)$ appears as a binding group relative to $\mathcal {Q}$ .

Choose a tuple $\bar {a}$ of elements $a_1=(\alpha _1,0),\ldots ,a_n=(\alpha _n,0)$ in the sort S with generic independent elements $\alpha _i$ in P and consider the set $G\star \bar a$ . By Proposition 3.11, the stationary type $\mathrm{stp}(\bar a/\ulcorner G\star \bar a\urcorner )$ is P-internal and fundamental with binding group G.

We now show that the binding group H of $\mathrm{stp}(\bar a/\ulcorner G\star \bar a\urcorner )$ relative to $\mathcal {Q}$ equals G. Assume for a contradiction that H is a proper subgroup of G. Since G is connected, the Morley rank $\mathrm{RM}(H)<\mathrm{RM}(G)$ , so

$$ \begin{align*} \mathrm{RM}(\bar{a}/ \ulcorner G\star\bar a\urcorner,\mathcal Q) < \mathrm{RM}(\bar{a}/ \ulcorner G\star\bar a\urcorner,P), \end{align*} $$

by [Reference Palacín and Pillay15, Fact 1.2]. Note that $\pi (\bar a)$ is definable over $\ulcorner G\star \bar a\urcorner $ . It follows that for some $1\leq k\leq n$ there is a tuple $\bar d$ with $\mathrm{stp}(\bar d)$ in $\mathcal Q$ such that the coordinate $a_k$ of $\bar a$ is not algebraic over $\bar {a}_{<k},\ulcorner G\star \bar a\urcorner $ , yet it is algebraic over $\bar {a}_{<k},\ulcorner G\star \bar a\urcorner ,\bar d$ .

By $\mathcal {P}$ -internality of $\mathrm{stp}(\bar d)$ , there is a set of parameters C with such that $\bar d$ is definable over $C,P$ . Therefore $a_k$ is algebraic over

$$ \begin{align*}\bar{a}_{<k},\ulcorner G\star\bar a\urcorner,C,P.\end{align*} $$

We deduce now from Lemma 3.8(c) that

$$ \begin{align*}\mathrm{stp}(\bar a/\ulcorner G\star\bar a\urcorner, C) \vdash \mathrm{tp}(\bar a/\ulcorner G\star\bar a\urcorner,C,P),\end{align*} $$

which contradicts that $a_k$ is not algebraic over $\bar {a}_{<k},\ulcorner G\star \bar a\urcorner $ , by the independence .

Proof Choose generic independent elements $a,b$ and c in the sort S and set $d=(a\otimes c)\oplus b$ . Consider now the following definable set:

$$ \begin{align*} E &= \big\{ (x,y)\in S^2 \ |\ \pi(x)=\pi(a) \ \& \ \pi(y)=\pi(b) \ \& \ d=(x\otimes c)\oplus y \big\}. \end{align*} $$

Since the canonical parameter $\ulcorner E\urcorner $ is clearly definable over $c,d,\pi (a),\pi (b)$ and the type $\mathrm{stp}(c,d,\pi (a),\pi (b)/\pi (c),\pi (d))$ is P-internal, we deduce that the type

$$ \begin{align*} \mathrm{stp}\big(\ulcorner E\urcorner/\pi(c),\pi(d)\big)\end{align*} $$

is P-internal.

Proof of the Claim Choose elements $a_1$ and $b_1$ in the fiber of $\pi (a)$ , resp. $\pi (b)$ , such that

Note that every automorphism $\sigma $ in $\mathrm{Aut}(\mathcal {M}_1/P)$ fixing the elements $a_1$ and $b_1$ must fix $\pi ^{-1}\big (\pi (a)\big )\times \pi ^{-1}\big (\pi (b)\big )\supseteq E$ , so $\sigma $ fixes E pointwise. In particular, the canonical parameter $ \ulcorner E\urcorner $ is definable over $a_1,b_1,P$ , as desired.

We assume now that $\mathcal M_1$ transfers internality to quotients in order to reach a contradiction. Since

$$ \begin{align*}\mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\pi(a),\pi(b)\big)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\pi(c),\pi(d)\big)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset),\end{align*} $$

we deduce that the type $\mathrm{stp}(\ulcorner E\urcorner )$ is almost P-internal. It follows that the type $\mathrm{stp}(b/a)$ is almost P-internal, since b is definable over $\ulcorner E\urcorner , a$ . Now, the independence yields the desired contradiction, because the sort S is not almost P-internal.

Proof Choose generic independent elements $a_1$ and $a_2$ in S and $\varepsilon $ in P generic over $a_1, a_2$ . Set $\bar \alpha =(\alpha _1,\alpha _2)=\big (\pi (a_1),\pi (a_2)\big )$ and let G be the definable subgroup of $(P^2,+)$ given by $\varepsilon \cdot x+y=0$ . Consider now the definable set $G\star (a_1,a_2)$ and choose $\beta _1$ in P generic over $ \ulcorner G\star (a_1,a_2)\urcorner , \bar \alpha ,\varepsilon $ as well as elements $\beta _2$ and $\beta _3$ in P with

(1) $$ \begin{align} 0&=\beta_1 \alpha_1 + \frac{1}{2}\beta_2 \alpha_{1}^2 + \frac{1}{3}\beta_3 \alpha_{1}^3+\alpha_2 \end{align} $$

(2) $$ \begin{align} 0&=\beta_1 +\beta_2 \alpha_1 + \beta_3 \alpha_{1}^2 - \varepsilon \end{align} $$

This is possible because the matrix

$$ \begin{align*} \begin{pmatrix} \frac{\alpha_1^2}{2} & \frac{\alpha_{1}^3}{3} \\ \alpha_1 & \alpha_{1}^2 \end{pmatrix}\end{align*} $$

has determinant $\frac {\alpha _{1}^4}{2}-\frac {\alpha _{1}^4}{3}\neq 0$ . Since $\beta _2$ and $\beta _3$ are definable over $\beta _1,\bar \alpha ,\varepsilon $ , we get the independence

(♦)

where $\bar \beta =(\beta _1,\beta _2,\beta _3)$ .

Proof of the Claim 1 Let $b_1,b_2$ and $b_3$ be elements in S such that $b_i$ is in the fiber of $\beta _i$ with

We show that every automorphism $\sigma $ in $\mathrm{Aut}(\mathcal {M}_{1}/P)$ fixing $b_1,b_2$ and $b_3$ must permute $G\star (a_1,a_2)$ . Recall that $F_\sigma $ is the derivation on $\mathbb {C}$ induced by the automorphism $\sigma $ . Since $F_{\sigma }(\beta _i)=0$ , we deduce from Equations (1) and (2) that $\varepsilon \cdot F_{\sigma }(\alpha _1)+F_{\sigma }(\alpha _2)=0$ . Since the tuple $\big (F_{\sigma }(\alpha _1),F_{\sigma }(\alpha _2)\big )$ lies in G, we have that

$$ \begin{align*} \sigma(a_1, a_2)= (F_{\sigma}(\alpha_1)\star a_1,F_{\sigma}(\alpha_2) \star a_2) \text{ belongs to } G\star (a_1,a_2), \end{align*} $$

so the automorphism $\sigma $ permutes the set $G\star (a_1,a_2)$ , by Proposition 3.11.

Proof of the Claim 2 Because of the independence $(\blacklozenge )$ , we need only show that

$$ \begin{align*}\mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\bar\beta\big)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}} \big(\bar\alpha,\varepsilon\big)=\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\emptyset).\end{align*} $$

Choose tuples $\bar {\beta }',\bar {\alpha }',\varepsilon ', \bar {\beta }",\bar {\alpha }",\varepsilon ", \bar {\beta }"'$ such that

$$ \begin{align*} \bar{\beta},\bar{\alpha},\varepsilon \stackrel{\text{st}}{\equiv} \bar{\beta}',\bar{\alpha},\varepsilon \stackrel{\text{st}}{\equiv} \bar{\beta}',\bar{\alpha}',\varepsilon' \stackrel{\text{st}}{\equiv} \bar{\beta}",\bar{\alpha}',\varepsilon' \stackrel{\text{st}}{\equiv} \bar{\beta}",\bar{\alpha}",\varepsilon" \stackrel{\text{st}}{\equiv} \bar{\beta}"',\bar{\alpha}",\varepsilon" \end{align*} $$

with

and

Since

$$ \begin{align*} \mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\bar{\beta}\big)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\bar{\alpha},\varepsilon\big)\subset \mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\bar{\beta}\big)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}\big(\bar{\beta}"'\big), \end{align*} $$

we need only show the independence

. Note first that the whole configuration has Morley rank 9:

$$ \begin{align*} \mathrm{RM}\big(\bar{\beta},\bar{\alpha},\varepsilon, \bar{\beta}',\bar{\alpha}',\varepsilon', \bar{\beta}",\bar{\alpha}",\varepsilon", \bar{\beta}"'\big) = \mathrm{RM}\big(\beta_1,\alpha_1,\alpha_2,\varepsilon, \beta_{1}',\alpha_{1}',\beta_{1}",\alpha_{1}",\beta_{1}"'\big) =9. \end{align*} $$

Since

$$ \begin{align*} \mathrm{RM}\big(\bar{\beta}"',\bar{\beta},\alpha_{1},\alpha_{1}',\alpha_{1}"\big)&= \mathrm{RM}\big( \bar{\beta}"'/\bar{\beta},\alpha_{1},\alpha_{1}',\alpha_{1}"\big) + \mathrm{RM}\big(\alpha_{1}" / \bar{\beta},\alpha_{1},\alpha_{1}' \big) \\ &\quad+ \mathrm{RM}\big( \alpha_{1}' / \bar{\beta},\alpha_{1} \big) + \mathrm{RM}\big(\alpha_{1} /\bar{\beta}\big) + \mathrm{RM}\big(\bar{\beta}\big) \\ &= \mathrm{RM}\big( \bar{\beta}"'/\bar{\beta},\alpha_{1},\alpha_{1}',\alpha_{1}"\big) + 6, \end{align*} $$

it suffices to show that $\alpha _{2},\varepsilon ,\bar {\beta }',\alpha _{2}',\varepsilon ', \bar {\beta }",\alpha _{2}"$ and $\varepsilon "$ are all algebraic over the tuple $(\bar {\beta }"',\bar {\beta },\alpha _{1},\alpha _{1}', \alpha _{1}")$ . Clearly $\alpha _{2},\varepsilon ,\alpha _{2}"$ and $\varepsilon "$ are algebraic over $\bar {\beta }"',\bar {\beta },\alpha _{1},\alpha _{1}"$ . Furthermore we have the following system of linear equations:

$$ \begin{align*} \begin{pmatrix} 6\alpha_{1} & 3\alpha_{1}^{2} & 2\alpha_{1}^{3} & 0 & 0 & 0 & 0 & 0 \\ 1 & \alpha_{1} & \alpha_{1}^{2} & 0 & 0 & 0 & 0 & 0 \\ 6\alpha_{1}' & 3{\alpha_{1}'}^{2} & 2{\alpha_{1}'}^{3} & 6 & 0 & 0 & 0 & 0 \\ 1 & \alpha_{1}' & {\alpha_{1}'}^{2} & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 & 6\alpha_{1}' & 3{\alpha_{1}'}^{2} & 2{\alpha_{1}'}^{3} \\ 0 & 0 & 0 & 0 & 1 & 1 & \alpha_{1}' & {\alpha_{1}'}^{2} \\ 0 & 0 & 0 & 0 & 0 & 6\alpha_{1}" & 3{\alpha_{1}"}^{2} & 2{\alpha_{1}"}^{3} \\ 0 & 0 & 0 & 0 & 0 & 1 & \alpha_{1}" & {\alpha_{1}"}^{2} \\ \end{pmatrix} \begin{pmatrix} \beta_{1}' \\ \beta_{2}' \\ \beta_{3}' \\ \alpha_{2}' \\ \varepsilon' \\ \beta_{1}" \\ \beta_{2}" \\ \beta_{3}" \end{pmatrix} = \begin{pmatrix} -6\alpha_{2} \\ \varepsilon \\ 0 \\ 0 \\ 0 \\ 0 \\ -6\alpha_{2}" \\ \varepsilon" \\ \end{pmatrix} \end{align*} $$

Thus, we need only show that the above matrix has non-zero determinant

$$ \begin{align*} &6 \left|\begin{matrix} 6\alpha_{1} \!\!& 3\alpha_{1}^2 \!\!& 2\alpha_{1}^3 \\ 1 \!\!& \alpha_{1} \!\!& \alpha_{1}^2 \\ 1\! \!& \alpha_{1}' \!\!& {\alpha_{1}'}^2 \end{matrix}\right| \left|\begin{matrix} 6\alpha_{1}' \!\!& 3{\alpha_{1}'}^2 \!\!& 2{\alpha_{1}'}^3 \\ 6\alpha_{1}" \!\!& 3{\alpha_{1}"}^2 \!\!& 2{\alpha_{1}"}^3 \\ 1 \!\!& \alpha_{1}" \!\!& {\alpha_{1}"}^2 \end{matrix}\right| -6 \left|\begin{matrix} 6\alpha_{1} \!\!& 3\alpha_{1}^2 \!\!& 2\alpha_{1}^3 \\ 1 \!\!& \alpha_{1} \!\!& \alpha_{1}^2 \\ 6\alpha_{1}' \!\!& 3{\alpha_{1}'}^2 \!\!& 2{\alpha_{1}'}^3 \end{matrix}\right| \left|\begin{matrix} 1 \!\!& \alpha_{1}' \!\!& {\alpha_{1}'}^2 \\ 6\alpha_{1}" \!\!& 3{\alpha_{1}"}^2 \!\!& 2{\alpha_{1}"}^3 \\ 1 \!\!& \alpha_{1}" \!\!& {\alpha_{1}"}^2 \end{matrix}\right| \\ &=72 \alpha_{1}^2 {\alpha_{1}'}^2 {\alpha_{1}"}^2 (\alpha_{1}-\alpha_{1}')(\alpha_{1}-\alpha_{1}") (\alpha_{1}"-\alpha_{1}')\neq 0. \\[-30pt] \end{align*} $$

If $\mathcal M_1$ had transfer of internality to intersections, then the type

$$ \begin{align*}\mathrm{stp}\big(\ulcorner G\star (a_1,a_2)\urcorner / \mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\ulcorner G\star (a_1,a_2)\urcorner)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(\beta_1,\beta_2,\beta_3)\big)\end{align*} $$

would be almost P-internal, by Claim 1, and so would be $\mathrm{stp}(\ulcorner G\star (a_1,a_2)\urcorner )$ , by the previous claim, which yields a contradiction, exactly as in the proof of Proposition 6.1.

Proof We may assume that the sort S is not almost P-internal. We want to show that the almost P-internal type $\mathrm{stp}(a,\gamma /B)$ is special, where a is a tuple $(a_1,\ldots ,a_n)$ of elements in S and $\gamma $ is a tuple in P. So, assume that for some set C of parameters the type $\mathrm{stp}(a,\gamma /C)$ is almost P-internal. By Lemma 3.10, the projections $\pi (a_i)$ are algebraic over B and over C. Hence, the type

$$ \begin{align*} \mathrm{stp}\big(a,\gamma/\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(B)\cap\mathrm{acl}^{\kern1.5pt \mathrm{eq}}(C)\big) \end{align*} $$

is almost P-internal and so is

since $\mathrm{Cb}(a,\gamma /B)$ is algebraic over finitely many B-conjugates of a.