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I have been trying to learn some stability theory lately, and I have been reading "Geometric Stability Theory" by Pillay. There is a lemma he states without proving which I am trying to prove, but I am getting stuck. The lemma states that if $b\in acl(aA)$ then $$R^{\infty}(tp(b/A))\leq R^\infty(tp(ab/A))=R^\infty(tp(a/A)).$$ Here is what I want to say: since $tp(ab/A)\models tp(a/A)$, we have $R^\infty(ab/A)\leq R^\infty(a/A)$. On the other hand, since $b\in acl(aA)$, it follows that $tp(a/A)\models tp(ab/A)$, and so $R^\infty(a/A)\leq R^\infty(ab/A)$, this takes care of the equality. But then since $tp(ab/A)\models tp(b/A)$, we must have $R^\infty(ab/A)\leq R^\infty(b/A)$, so that if the inequality were true, we would necessarily have equality throughout, which I find rather unlikely. So I am pretty sure that there is something wrong with my reasoning, but I am not sure exactly what it is.

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This question was asked over a month ago, but I'll answer it in case you are still interested (or some future person is).

There is a tempting mistake in your argument (which I also made when I thought about this for the first time): The fact you are trying to use is that for (partial) types $p(x)$ and $q(x)$ in the same tuple of variables $x$, $p\rightarrow q$ implies $R^\infty(p)\leq R^\infty(q)$.

This does not hold if $p$ and $q$ have different variables. As a simple example of the impact of changing variables, let $a$ be an element of an infinite set. Let $\phi(x,y)$ be the formula $x = a$ and let $\psi(x)$ also be $x = a$ thought of in a context with just $x$. Then $\phi\rightarrow \psi$, but $R^\infty(\psi) = 0$ (since $\psi$ is algebraic), while $R^\infty(\phi)>0$ (since $\phi$ is not), i.e. $\phi$ picks out an entire line in the plane, while $\psi$ only picks out a point.

To prove Pillay's lemma, try proving the following two statements by transfinite induction. The assumptions are crafted to let you bridge the differences in variable contexts.

  1. If $\Phi(x)$ and $\Psi(x,y)$ are partial types over $A$ such that $\Phi(x)\rightarrow \exists y \Psi(x,y)$, then $R^\infty(\Phi(x)) \leq R^\infty(\Psi(x,y))$. Here $\Psi$ may be an infinite collection of formulas, in which case $\exists y \Psi(x,y)$ means "slap an existential quantifier in front of all finite conjunctions of formulas from $\Psi$."

  2. Let $\theta(x,y)$ be an $\mathcal{L}(A)$-formula, and let $\theta^*(x,y)$ be the formula $\theta(x,y)\land \exists_1^nx\theta(x,y)$, where $\exists_1^n$ expresses ``there exists at least $1$ and at most $n$". If $\Phi(x,y)$ and $\Psi(y)$ are partial types over $A$ such that $\Phi(x,y)\rightarrow \Psi(y)\cup\{\theta^*(x,y)\}$, then $R^\infty(\Phi(x,y))\leq R^\infty(\Psi(y))$.

From the first, you can conclude that $R^\infty(\text{tp}(b/A))\leq R^\infty(\text{tp}(ab/A))$ and $R^\infty(\text{tp}(a/A))\leq R^\infty(\text{tp}(ab/A))$ in general (without using algebraicity).

From the second, letting $\theta$ isolate $\text{tp}(b/Aa)$, you can conclude that $R^\infty(\text{tp}(ab/A))\leq R^\infty(\text{tp}(a/A))$

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