Imaginaries 1
Imaginaries after Poizat («Une théorie de Galois imaginaire».
- A theory is said to weakly eliminate imaginaries if every formula ${\varphi}$ has a minmal algebraically closed set of parameters (i.e. a set ${A=\mathrm{acl}(A)}$ such that ${\varphi(x,b) = \psi(x,a)}$ with ${a\in A}$).
- A theory is said to eliminate imaginaries if every formula ${\varphi}$ has the (definably closed) canonical set of parameters or canonical base, i.e. any automorphism preserves ${\varphi}$ set-wise iff it preserves ${B=\mathrm{dcl}(B)}$ point-wise.
Clearer approach to imaginaries.
- A theory ${T}$ is said to weakly eliminate imaginaries if for every ${a \in M^\mathrm{eq}}$ there exists ${b \in M^n}$ such that ${b \in \mathrm{acl}(a)}$, ${a \in \mathrm{dcl}(M)}$
- A theory ${T}$ is said to eliminate imaginaries if for every ${a \in M^\mathrm{eq}}$ there exists ${b \in M^n}$ such that ${b \in \mathrm{dcl}(a)}$, ${a \in \mathrm{dcl}(M)}$
The first is equivalent to the second (proved in Poizat).
Strong elimination of imaginaries is the weak elimination of imaginaries plus elimination of finite imaginaries, i.e. those such that the equivalence classes are finite.
Notable example of weak elimination:
Theorem 1 Let ${T}$ be a strongly minimal theory with infinite ${\mathrm{acl}{}(\emptyset)}$. Then ${T}$ weakly eliminates imaginaries.
Proof: Let ${e \in M^{eq}}$ be the image of a 0-definable function, ${f(\bar{b})=e}$, ${\bar{b} \in M^n}$. We need to prove that there are only finitely many such ${\bar{c}}$ that ${f(\bar{c})=e}$. It boils down to finding one such ${\bar{c}}$ in ${\mathrm{acl}{}(e)}$. Suppose we have already found such ${c_0}$ that ${\exists x_1\ldots x_n f(c_0,x_1,\ldots,x_n)=e}$. Then let ${g(x_1)=\exists x_2\ldots x_n f(c_0,x_1,x_2,\ldots,x_n)}$. Since ${T}$ is strongly minimal, ${g}$ defines either a finite or cofinite set in ${M}$. If it is finite, then it belongs to ${\mathrm{acl}{}(e)}$ and we take on of its elements as ${c_1}$. Otherwise, it is cofinite and it intersects with ${\mathrm{acl}{}(M)}$, because the latter is infinite, and we can pick a ${c_1}$ from the intersection. By iterating this process, we get ${\bar{c}}$. ☐
Thus, ACF weakly eliminates imaginaries.
Theorem 2 ACF eliminates finite imaginaries as well.
Proof: Symmetric polynomials ${p_1(\bar{x}), \ldots, p_k(\bar{x})}$ are by definition the generators of ${k[x_1,\ldots,x_n]^{\Sigma_n}}$. It follows that ${p_i(\bar{a})=p_i(\bar{b})}$ for all ${i}$ iff ${\bar{a}}$ is a permutation of ${\bar{b}}$.
A finite imaginary is 0-interdefinable with an imaginary for the equivalence relation ${(\bar{x}_1, \dots, \bar{x}_n) \sim_E (\bar{y}_1, \dots, \bar{y}_n)}$ where ${\bar{y}_i=\bar{x}_{\sigma(i)}}$ for some permutation ${\sigma}$.
The function ${f(\bar{z})=\langle p_i(\ \prod_j (z-x_{1j}),\ldots, \prod_j (z-x_{nj})\ ) \rangle}$ eliminates such an imaginary. ☐