I was glancing at this paper of Chatzidakis and became a bit confused by the definitions of (weak) elimination of imaginaries given in it. Fix a complete theory $T$ and a monster model $\mathfrak{U}$; recall that $T$ has elimination of imaginaries (EI) if every $\mathfrak{U}$-definable subset $D\subseteq\mathfrak{U}^r$ has a canonical parameter – ie a finite tuple $\overline{c}:=\ulcorner D\urcorner$ that is fixed pointwise by precisely the automorphisms of $\mathfrak{U}$ that fix $D$ setwise. Likewise, $T$ has weak elimination of imaginaries (WEI) if every $\mathfrak{U}$-definable subset $D\subseteq\mathfrak{U}^r$ is definable over a smallest algebraically closed set. Since the intersection of a family of algebraically closed sets is certainly algebraically closed, this holds if and only if, whenever $(A_i)_{i\in I}$ is a family of algebraically closed sets over each of which $D$ is definable, $D$ is also definable over $\bigcap_{i\in I}A_i$. Now, Section 2.37 of the linked paper claims that EI (resp. WEI) is equivalent to condition 1 (resp. condition 2) below:
- If $D\subseteq \mathfrak{U}^r$ is a $\mathfrak{U}$-definable set, and $D$ is definable over a tuple $\overline{a}$ and over a tuple $\overline{b}$, then $D$ is definable over $\operatorname{dcl}(\overline{a})\cap\operatorname{dcl}(\overline{b})$.
- If $D\subseteq \mathfrak{U}^r$ is a $\mathfrak{U}$-definable set, and $D$ is definable over a tuple $\overline{a}$ and over a tuple $\overline{b}$, then $D$ is definable over $\operatorname{acl}(\overline{a})\cap\operatorname{acl}(\overline{b})$.
Unfortunately, I'm struggling to see why these definitions work. My first question is about condition 1, which I think I certainly must be misunderstanding. For instance, if $T$ is the theory of infinite sets, it is fairly easy to see that $T$ has WEI but not EI. But $\operatorname{acl}$ and $\operatorname{dcl}$ coincide in $T$ by quantifier elimination, so aren't conditions 1 and 2 equivalent in that context?
My second question is about condition 2, which I'm struggling to show implies WEI. Condition 2 of course tells us that, if $(A_i)_{i\in I}$ is any finite family of algebraically closed sets over which $D$ is definable, then $D$ is also definable over $\bigcap_{i\in I}A_i$. But it's not clear to me how we can extend this to when $I$ is infinite; I suspect we can probably use some compactness argument, but it's not coming to my mind. The problem seems to me that the property of "being algebraic" over a tuple is not in general first-order, and indeed in general not even type-definable, so I'm not sure how to proceed. Any insight or ideas would be appreciated!
Alex's answer below has completely resolved the questions above, and more! However, for posterity, here's an example showing that at least one of the equivalences he's proved does not in general hold when $D$ above is replaced by a type-definable set. In particular, this example shows that the natural analogue of condition 1 above for a type-definable set $D$ is not in general equivalent to the statement that $D$ is type-definable over a smallest $\operatorname{dcl}$-closed set.
Consider a multi-sorted theory $T$ with sorts $S_n$ for all $n\in\mathbb{N}$. Let the language include a function $f_n:S_n\to S_{n+1}$ for each $n\in\mathbb{N}$, and let the axioms of $T$ state that each $f_n$ is a three-to-one function, ie that the fiber of $f_n$ over every point of $S_{n+1}$ has size precisely $3$. If I'm not mistaken, $T$ is $\aleph_0$-categorical with no finite models, and hence complete; let $\mathfrak{U}$ be a monster model, and let $U_n$ be the realization of $S_n$ in $\mathfrak{U}$ for each $n\in\mathbb{N}$. Note that, for any $a\in U_n$, we have $$\operatorname{dcl}(a)=\{a\}\cup\{(f_{n+k}\circ\dots\circ f_n)(a):k\in\mathbb{N}\},$$ since the fibers of each $f_n$ have size $3$. Now, for each $n,k\in\mathbb{N}$, define the formula $$\varphi^k_n(v,w)\equiv (f_{n+k}\circ\dots\circ f_n)(v)=(f_{n+k}\circ\dots\circ f_n)(w)$$ on $S_n$; note that, for any $a\in U_n$, we have $\# \varphi^k_n(a,U_n)=3^{k+1}$. Given any $a\in U_n$, let $F_n(a)=\bigcup_{k\in\mathbb{N}}\varphi_n^k(a,U_n)$. Note that, for any $a,b\in U_n$, there is an automorphism of $\mathfrak{U}$ swapping $F_n(a)$ and $F_n(b)$ and fixing everything else in $U_n$.
Now, fix any $a\in U_0$, and consider the partial type $p(v)=\{\neg\varphi_0^k(a,v):k\in\mathbb{N}\}$ on $S_0$; I claim that $p(U_0)=U_0\setminus F_0(a)$ satisfies (the analogue of) condition 1, but is not type-definable over a smallest definably closed set. To see that $p$ satisfies condition 1, note first that $p(U_0)$ is type-definable over some $e\in U_n$ if and only if $e\in (f_n\circ\dots\circ f_0)\left(F_0(a)\right)$; indeed, let $e'\in U_0$ be any element such that $e=(f_n\circ\dots\circ f_0)(e')$. Then if $e'\in F_0(a)$, we have $F_0(e')=F_0(a)$, so certainly $p$ is definable over $e'$ (and hence over $e$). If $e'\notin F_0(a)$, then $F_0(e')\cap F_0(a)=\varnothing$; letting $b\in U_0$ be any element such that $F_0(b)\cap\left[F_0(a)\cup F_0(e')\right]=\varnothing$, by the remark above there is an automorphism of $\mathfrak{U}$ fixing $e'$, hence $e$, and swapping $F_0(a)$ and $F_0(b)$, hence not fixing $p(U_0)$ setwise. Condition 1 follows; suppose $p(U_0)$ is type-definable over some $d\in U_m$ and some $e\in U_n$. Then we have $d\in (f_m\circ\dots\circ f_0)\left(F_0(a)\right)$ and $e\in (f_n\circ\dots\circ f_0)\left(F_0(a)\right)$, so there exists some $k\in\mathbb{N}$ with $k\geqslant m,n$ such that $$(f_{k}\circ\dots\circ f_{m+1}\circ f_m)(d)=(f_k\circ\dots\circ f_0)(a)=(f_{k}\circ\dots\circ f_{n+1}\circ f_n)(e).$$ Then $p(U_0)$ is type-definable over this latter element, which lies in $\operatorname{dcl}(d)\cap\operatorname{dcl}(e)$, and so we are done.
However, $p(U_0)$ is not type-definable over a smallest $\operatorname{dcl}$-closed set. To see this, for each $k\in\mathbb{N}$ define $$X_k=\operatorname{dcl}\left((f_{k}\circ\dots\circ f_0)(a)\right)=\{(f_n\circ\dots\circ f_0)(a):n\geqslant k\}.$$ By the remarks above $p(U_0)$ is type-definable over each $X_k$, and each $X_k$ is $\operatorname{dcl}$-closed. But $\bigcap_{k\in\mathbb{N}}X_k=\varnothing$, over which $p(U_0)$ is certainly not type-definable, so we're done.
Best Answer
$\newcommand{\dcl}{\mathrm{dcl}} \newcommand{\acl}{\mathrm{acl}} \newcommand{\eq}{\mathrm{eq}} \newcommand{\tp}{\mathrm{tp}}$
You're exactly right: Chatzidakis's condition 1 is not equivalent to EI. Condition 2 is equivalent to WEI, but the equivalence is not obvious (at least it wasn't obvious to me).
I'll write DCLF for the property that for every definable set $D$, there is a smallest $\dcl$-closed set over which $D$ is definable. Similarly, I'll write ACLF for the property that for every definable set $D$, there is a smallest $\acl$-closed set over which $D$ is definable. I'll show below that Chatzidakis's conditions 1 and 2 are equivalent to DCLF and ACLF, respectively.
I've borrowed the name DCLF from the paper Weak forms of elimination of imaginaries by Casanovas and Farré. This paper is a wonderful reference for all the various properties related to EI. Casanovas and Farré discuss DCLF on p. 132 and give the theory of infinite sets as a counterexample to the equivalence with EI, just as you did. Proposition 2.10 clarifies the exact sense in which DCLF is weaker than EI: DCLF plus coding finite sets of pairwise interdefinable tuples is equivalent to EI (compare with the fact that WEI plus coding arbitrary finite sets of tuples is equivalent to EI). Finally, Casanovas and Farré point out on p. 128 that Poizat's original paper on EI, Une Théorie de Galois Imaginaire, mistakenly claimed that EI and DCLF were equivalent, and this error has propagated to some other sources - I would guess that this is the origin of the mistake in the Chatzidakis notes.
On the other hand, ACLF is well-known to be equivalent to WEI (and it is sometimes taken as the definition of WEI, as you did in your question). But the fact that DCLF is not equivalent to EI means that I would prefer not to take ACLF as the definition of WEI. The usual definition is: for every imaginary element $e\in M^{\eq}$, there is a real tuple $a\in M$ such that $e\in \dcl^{\eq}(a)$ and $a\in \acl^{\eq}(e)$. Or if you prefer to give an "intrinsic" definition not referring to imaginaries: for every definable set $D$, there is a finite set $F$ of tuples such that an automorphism fixes $D$ setwise if and only if it fixes $F$ setwise. Both of these definitions are natural generalizations of conditions that do correctly characterize EI (replace $\acl^{\eq}$ with $\dcl^{\eq}$ in the first, and replace the finite set $F$ with a single tuple in the second).
Ok, enough discussion, on to the proofs. In the following arguments, I'll have to bounce back and forth between $M$ and $M^{\eq}$, so just to be clear about notation: $\acl^{\eq}$ is algebraic closure computed in $M^{\eq}$. When $B\subseteq M^{\eq}$, I'll write $B\cap M$ for the "real part" of $B$. When $A\subseteq M$ is real, $\acl(A)$ is the algebraic closure of $A$ computed in $M$, and note that $\acl(A) = \acl^{\eq}(A)\cap M$. The same goes for $\dcl$ and $\dcl^{\eq}$.
First, it is clear that DCLF implies condition 1 and ACLF implies condition 2. If there is a smallest $\dcl/\acl$-closed set $C$ over which $D$ is definable, then for any two $\dcl/\acl$-closed sets $A$ and $B$ over which $D$ is definable, $C$ is contained in $A\cap B$, so $D$ is definable over the $\dcl/\acl$-closed set $A\cap B$.
Conversely, suppose $T$ satisfies condition 2. We'll show it satisfies ACLF. Let $D$ be a definable set, and let $e\in M^{\eq}$ be its code. Let $\varphi(x,a)$ be any formula defining $D$ with $a\in M$. Then $e\in \dcl^{\eq}(a)$. By extension for algebraic independence (see here, for example) in $M^{\eq}$, we can find $b\in M$ with $\tp(b/e) = \tp(a/e)$ such that $\acl^{\eq}(ae)\cap \acl^{\eq}(be) = \acl^{\eq}(e)$. Since $\tp(b/e) = \tp(a/e)$, $e\in \dcl^{\eq}(b)$, so $D$ is definable over $b$. Then by condition 2, $D$ is definable over the $\acl$-closed set $C = \acl(a)\cap \acl(b)$. If $D$ is definable over some other $\acl$-closed set $C'\subseteq M$, then $e\in \dcl^{\eq}(C')$, so $$C\subseteq (\acl^{\eq}(a)\cap \acl^{\eq}(b))\cap M \subseteq \acl^{\eq}(e)\cap M \subseteq \acl^{\eq}(C')\cap M = \acl(C') = C'.$$ Thus $C$ is the smallest $\acl$-closed set over which $D$ is definable.
Finally, suppose $T$ satisfies condition 1. We'll show it satisfies DCLF. Let $D$ be a definable set, and let $e\in M^{\eq}$ be its code. Since condition 1 is stronger than condition 2, the proof above shows that $D$ is definable over $\acl^{\eq}(e) \cap M$. For each finite tuple $a$ in $\acl^{\eq}(e) \cap M$, let $N_a$ be the number of realizations of $\tp(a/e)$. Pick some finite tuple $c\in \acl^{\eq}(e) \cap M$ such that $D$ is definable over $c$ and such that $N_c$ is minimal. I claim that $C = \dcl(c)$ is the smallest $\dcl$-closed set over which $D$ is definable.
First we'll show that $C$ is a minimal $\dcl$-closed set over which $D$ is definable. Let $C'\subseteq C$ be a $\dcl$-closed set such that $D$ is definable over $C'$. Let $c'\in C'$ be a finite tuple over which $D$ is definable. Since $c'\in C = \dcl(c)$, $N_{c'}\leq N_c$. But by minimality of $N_c$, $N_{c'} = N_c$. It follows that $c\in \dcl(c')$. But then $C = \dcl(c) \subseteq \dcl(c')\subseteq C'$.
Now let $B$ be any $\dcl$-closed set over which $D$ is definable, and let $b\in B$ be a finite tuple over which $D$ is definable. By condition 1, $D$ is definable over $\dcl(b)\cap \dcl(c)\subseteq B\cap C \subseteq C$. By minimality, $C = B\cap C$, so $C\subseteq B$. Thus $C$ is the smallest $\dcl$-closed set over which $D$ is definable.