Let $T$ be a stable theory, and $M$ a sufficiently saturated model. Let $x,b,E $ such that $x \in \operatorname{acl}(MEb)$ and such that $x\overset{\vert}{\smile}_{Mb}E$. Then $x \in \operatorname{acl}(Mb)$.

On the intuitive level, this is very clear to me, it just says that the $E$ part making $x$ algebraic is unnecessary (by independence), however I can't seem to prove this using the ''axiomatic'' properties of the independence relation $\overset{\vert}{\smile}_{}$ i.e symmety, transitivity, etc. I also tried arguing by contradiction, using the exchange property for $\operatorname{acl}$.

I stumbled upon this small step while working through a proof of the group configuration for stable theories, but I don't think the context matters much.


3 Answers 3


Since you're happy to assume $T$ is stable, the other answers have surely settled your question. But the fact is true in an arbitrary theory $T$, so I'd like to give you an additional proof that does not rely on stability.

First a definition. We say $A$ and $B$ are algebraically independent over $C$, written $\newcommand{\acl}{\mathrm{acl}} \newcommand{\tp}{\mathrm{tp}} \newcommand{\ind}{\overset{\vert}{\smile}} A\ind^a_CB$, if $\acl(AC)\cap \acl(BC) = \acl(C)$. To avoid confusion, I'll write $\ind^f$ for forking independence.

Lemma (Full existence for algebraic independence): Let $A$, $B$, $C$ be arbitrary sets. Then there exists $A'$ with $\tp(A'/C) = \tp(A/C)$ and $A'\ind^a_C B$.

The lemma follows from "Strong elementary amalgamation over algebraically closed sets", Theorem 6.4.5 in Hodges Model Theory or Theorem 5.3.5 in Hodges A Shorter Model Theory.

Now suppose $a\ind^f_B E$ and $a \in \acl(BE)$. Then there is some algebraic $L(B)$-formula $\varphi(x,e)\in \tp(a/BE)$, and we may assume that $\varphi(x,e)$ isolates $\tp(a/BE)$. By the lemma, we can find a sequence $(e_n)_{n\in \omega}$ such that $e_0 = e$, $\tp(e_n/B) = \tp(e/B)$, and $e_n\ind^a_B e_0\dots e_{n-1}$ for all $n$. Since $a\ind^f_BE$, $\varphi(x,e)$ does not divide over $B$. So $\{\varphi(x,e_n)\mid n\in \omega\}$ is not $2$-inconsistent. In particular, there exist $n< n'$ and $a'$ such that $\varphi(a',e_n)$ and $\varphi(a',e_{n'})$. But then $a'\in \acl(Be_{n'})\cap \acl(Be_{n}) = \acl(B)$, since $e_{n'}\ind^a_B e_n$. Finally, since $\tp(a'e_n/B) = \tp(ae/B)$, there is an automorphism of the monster model fixing $B$ and moving $a'e_n$ to $ae$. This automorphism fixes $\acl(B)$ setwise, so $a\in \acl(B)$.

  • $\begingroup$ I was actually missing this "algebraic independence" notion, it ties this up very nicely. Thanks! $\endgroup$ Apr 29, 2021 at 12:12

Once you have certain properties of independence in stable theories, we can do this with just those properties. The proof of one of these properties is essentially the answer from Atticus, so ultimately you will always need an argument like that. As in their answer we can also prove the stronger statement that $a \overset{\vert}{\smile}_B E$ and $a \in \operatorname{acl}(BE)$ implies $a \in \operatorname{acl}(B)$ for any $a, B, E$.

The properties we need are:

  1. We have $a \overset{\vert}{\smile}_B a$ if and only if $a \in \operatorname{acl}(B)$, for any $a$ and $B$.
  2. Normality: $a \overset{\vert}{\smile}_B E$ implies $a \overset{\vert}{\smile}_B BE$ for any $a, B, E$.
  3. Transitivity: $a \overset{\vert}{\smile}_B C$ and $a \overset{\vert}{\smile}_C D$ implies $a \overset{\vert}{\smile}_B D$ for any $B \subseteq C$ and $D$.

The first property is a simpler version of what Atticus proved in their answer. Normality holds in any theory (for forking/dividing independence) and transitivity holds for dividing independence if and only if the theory is simple, so in particular it holds in stable theories.

Now suppose $a \overset{\vert}{\smile}_B E$ and $a \in \operatorname{acl}(BE)$. Then by the first property $a \overset{\vert}{\smile}_{BE} a$. From $a \overset{\vert}{\smile}_B E$ we get by normality that $a \overset{\vert}{\smile}_B BE$. We can now apply transitivity to find $a \overset{\vert}{\smile}_B a$. By the first property again we then conclude $a \in \operatorname{acl}(B)$, as required.

  • $\begingroup$ oh nice, this is much cleaner! (+1) $\endgroup$ Apr 28, 2021 at 13:40
  • $\begingroup$ Thank you Mark, I'll mark this as the answer since it is exactly what I was lookng for. $\endgroup$ Apr 29, 2021 at 12:13

In fact, we don't need $M$ to be a model here; the result holds with $Mb$ replaced by an arbitrary parameter set $B$.

Suppose $a\perp_BE$ and $a\notin\operatorname{acl}(B)$; we will show $a\notin\operatorname{acl}(BE)$. It suffices to show that $a\notin\operatorname{acl}(Be)$ for any (finite) tuple $e\in E$, so let $e$ be such, and let $\pi(v,w)=\operatorname{tp}(a,e/B)$. Now, since $T$ is stable, independence is symmetric, and so we have $e\perp_{B}a$. Also, since $a\notin\operatorname{acl}(B)$, let $\mathcal{I}=(a_i)_{i\in\omega}$ be an infinite family of distinct $B$-conjugates of $a$. By the "standard lemma", we may assume that $\mathcal{I}$ is $B$-indiscernible, and then (since $e\perp_{B}a$), we have that $\bigcup_{i\in\omega}\pi(a_i,w)$ is consistent. Let $e'$ be a realization. Then $a_i\equiv_{Be'} a_j$ for all $i,j$, so in particular $a_0\notin\operatorname{acl}(Be')$. But $a_0e'\equiv_{B}ae$, so this means $a\notin\operatorname{acl}(Be)$ as well, as desired.


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