Let T be a totally transcendental theory. We have that $\text{RM}(a/A,b)=\text{RM}(a/A)$ if and only if $\text{RM}(b/A,a)=\text{RM}(b/A)$. Doesn't it follow from this that acl defines a pregeometry on sets with Morley rank one? (I do only know that this is true for strongly minimal sets.) Sketch of proof: $b\notin\text{acl}(A,a)$ implies $\text{RM}(b/A,a)=1=\text{RM}(b/A)$. Hence, $\text{RM}(a/A,b)=\text{RM}(a/A)$. If $a\notin\text{acl}(A)$, we have $\text{RM}(a/A)=1$ and therefore $\text{RM}(a/A,b)=1$. Thus $a\notin\text{acl}(A,b)$.

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    $\begingroup$ Why the downvotes on this reasonable question by a new user? $\endgroup$ Commented Jun 30, 2018 at 14:46

1 Answer 1


Your argument works just fine, and algebraic closure does indeed define a pregeometry on sets of Morley rank $1$.

But this is not so surprising: sets of Morley rank $1$ are not so far away from strongly minimal sets. That is, a set is strongly minimal if and only if it has Morley rank $1$ and Morley degree $1$. A general set of Morley rank $1$ has some Morley degree $d$, and it can be expressed as a union of $d$ strongly minimal sets (its "irreducible components").

So why do we focus on strongly minimal sets, when we just as well get pregeometries from sets of Morley rank $1$? Because a strongly minimal set $D$ has the convenient extra property that over any closed subset $X$, any two elements not in $X$ have the same type over $X$ and the parameters defining $D$. It follows that (assuming we're working in a saturated model, to be safe) they are conjugate by an automorphism of the model fixing $D$ setwise and $X$ pointwise. This means that the pregeometry is homogeneous.

In a general set of Morley rank $1$, this isn't true, since to specify the type of $a$ over $X$, we might have to specify which of the $d$ irreducible components $a$ lives in.

  • $\begingroup$ I thank you very much for this answer. $\endgroup$
    – paul
    Commented Jun 30, 2018 at 15:31

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