Our proof of model completeness for ACFA yields\u00a0 strong quantifier simplification results.<\/p>\n
Let us start with a relative completeness result.\u00a0 \u00a0If a theory $T$ is model complete, then whenever $E \\models T$ and $E \\subseteq M_i \\models T$ for $i = 1$ and $2$, then $M_1 \\equiv_E M_2$.\u00a0 For ACFA we can replace the condition $E \\models T$ with just that $E$ is a substructure which is algebraically closed as a field.<\/p>\n
Proposition:<\/strong>\u00a0 Let $(E,\\sigma)$ be a difference field which is algebraically closed as a field.\u00a0 Suppose that $(E,\\sigma) \\subseteq (M_i,\\sigma_i) \\models \\text{ACFA}$ are two extensions to existentially closed difference fields for $i = 1$ and $2$, then $M_1 \\equiv_E M_2$.<\/p>\n Proof:<\/strong>\u00a0 Since $E$ is algebraically closed as a field, the ring $R := M_1 \\otimes_E M_2$ is an integral domain (see Corollary 2 on page 202 of Nathan Jacobson, Lectures in Abstract Algebra III. Theory of Fields and Galois Theory<\/a>, GTM 32.)\u00a0 and has a distinguished endomorphism $\\sigma$ defined on tensors via $a \\times b \\mapsto \\sigma_1(a) \\otimes \\sigma_2(b)$.\u00a0 \u00a0 Let $K := \\mathcal{Q}(R)$ be the field of fractions of $R$ with $\\sigma:K \\to K$ the extension of $\\sigma$ on $R$ defined by $\\sigma(\\frac{a}{b}) = \\frac{\\sigma(a)}{\\sigma(b)}$.\u00a0 Let $(L,\\sigma) \\models \\text{ACFA}$ be an existentially closed extension of $(K,\\sigma)$.\u00a0 Then because ACFA is model complete, we have two elementary exrtenions $M_1 \\preceq L$ and $M_2 \\preceq L$.\u00a0 As $E$ is a common substructure of both $M_1$ and $M_2$, a fortiori<\/em>, we have $M_1 \\equiv_E L \\equiv_E M_2$.\u00a0 \u00a0$\\Box$<\/p>\n Corollary:<\/strong>\u00a0 The completions of ACFA are in correspondence with pairs $(p,[\\sigma])$ where $p$ is a prime number or $0$ and $[\\sigma]$ is a conjugacy class in $\\operatorname{Gal}(k_0^\\text{alg}\/k_0)$ with $k_0 = \\mathbb{F}_p$ if $p$ is prime and\u00a0 $k_0 = \\mathbb{Q}$ if $p = 0$.<\/p>\n Proof:\u00a0 \u00a0<\/strong>For $(K,\\sigma) \\models \\text{ACFA}$, let $\\chi((K,\\sigma)) := (p,[\\sigma])$ where $p$ is the characteristic of $K$ and $[\\sigma]$ is the conjugacy class of $\\sigma \\upharpoonright_{k_0^\\text{alg}}$ with $k_0$ the prime field in $K$.\u00a0 As $K$ is algebraically closed, $k_0^\\text{alg} \\subseteq K$.<\/p>\n If $(K_i,\\sigma_i) \\models \\text{ACFA}$ are models of ACFA for $i = 1$ and $2$ and $\\chi((K_1,\\sigma_1)) = \\chi((K_2,\\sigma_2))$, then $K_1$ and $K_2$ have a common prime field $k_0$ (as they have the same characteristic) and there is some $\\tau \\in \\operatorname{Gal}(k_0^\\text{alg}\/k_0)$ with $\\sigma_2 \\upharpoonright_{k_0^\\text{alg}}\u00a0 = \\tau (\\sigma_2 \\upharpoonright_{k_0^\\text{alg}} ) \\tau^{-1}$.\u00a0 \u00a0That is, $\\tau:(k_0^\\text{alg},\\sigma_1 \\upharpoonright_{k_0^\\text{alg}} ) \\to (k_0^\\text{alg},\\sigma_2 \\upharpoonright_{k_0^\\text{alg}} )$ is an isomorphism of difference fields.\u00a0 Set $E := (k_0^\\text{alg},\\sigma_1 \\upharpoonright_{k_0^\\text{alg}} )$.\u00a0 Then via the usual inclusion, $E$ is an algebraically closed substructure of $K_1$ and via $\\tau$, it is an algebraically closed substructure of $K_2$.\u00a0 By the above proposition, $K_1 \\equiv_E K_2$, which implies that upon forgetting $E$, we have $K_1 \\equiv K_2$.<\/p>\n On the other hand, if $K_1$ and $K_2$ are models of ACFA with $K_1 \\equiv K_2$, then clearly they have the same characteristic, and thus the same prime field $k_0$.\u00a0 As they are both algebraically closed, they both contain (copies of) $k_0^\\text{alg}$.\u00a0 Thus to see that $\\chi(K_1) = \\chi(K_2)$ it suffices to show that for each finite Galois extension $L\/k_0$ of $k_0$ there is an isomorphism $\\tau:(L,\\sigma_1 \\upharpoonright_L) \\to (L,\\sigma_2 \\upharpoonright_L)$, where here “$\\sigma_i \\upharpoonright_L$” means the restriction $\\sigma_i$ to $L$ under some embedding of $L$ into $K_i$ for $i = 1$ and $2$.\u00a0 \u00a0By the primitive element theorem, we may express $L = k_0[a]$ for some element $a \\in L$.\u00a0 Let $P(x) \\in k_0[x]$ be the minimal monic polynomial of $a$ over $k_0$.\u00a0 Then\u00a0 is a polynomial $Q \\in k_0[x]$ so that $\\sigma_1(a) = Q(a)$.\u00a0 As $K_1 \\equiv K_2$, $K_2 \\models (\\exists b) (P(b) = 0 \\land \\sigma(b) = Q(b))$.\u00a0 Define $\\tau:(L,\\sigma_1 \\upharpoonright_L) \\to (K_2,\\sigma)$ by $a \\mapsto b$.\u00a0 \u00a0A general element of $L$ may be expressed as $R(a)$ for some $R \\in k_0[x]$.\u00a0 We compute that $\\tau \\sigma_1 (R(a)) = \\tau (R(\\sigma(a)) = \\tau R(Q(a)) = R(Q(b)) = R(\\sigma_2(b)) = \\sigma_2(R(b))$.\u00a0 Thus, $\\tau$ is an isomorphism of difference fields and in the direct limit we have an isomorphism $\\tau:(k_0^\\text{alg},\\sigma_1 \\upharpoonright_{k_0^\\text{alg}}) \\to (k_0^\\text{alg},\\sigma_2 \\upharpoonright_{k_0^\\text{alg}})$.\u00a0 That is, $\\sigma_2 \\upharpoonright_{k_0^\\text{alg}} = \\tau \\circ \\sigma_1 \\upharpoonright_{k_0^\\text{alg}} \\circ \\tau^{-1}$.\u00a0 $\\Box$<\/p>\n Remark:<\/strong>\u00a0 It follows from the fact that Frobenius automorphism topologically generates $\\operatorname{Gal}(\\mathbb{F}_p^\\text{alg}\/\\mathbb{F}_p)$ in characteristic $p$ and the Chebotarev density theorem in characteristic zero that for $k_0$ the prime field every difference field of the form $(k_0^\\text{alg},\\sigma)$ is isomorphic to the the algebraic closure of the prime field in $\\prod_\\mathcal{U} (\\mathbb{F}_q^\\text{alg},x \\mapsto x^q)$ where $\\mathcal{U}$ is a nonprincipal ultrafilter on the set of prime powers.\u00a0 In fact, in characteristic zero, we may assume that $\\mathcal{U}$ is concentrated on the primes.<\/p>\n Proposition:\u00a0<\/strong> If $(K,\\sigma) \\models \\text{ACFA}$, $E \\subseteq K$ is an algebraically closed substructure, and $a$ and $b$ are two tuples from $K$ of the same length, then $\\operatorname{tp}(a\/E) = \\operatorname{tp}(b\/E)$ if and only if there is an isomorphism of difference fields over $E$ between $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ and $E \\langle b \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ taking $a$ to $b$ where the notation $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}$ means the inversive difference field generated by $a$ over $E$, that is, the field $E( \\{ \\sigma^j (a) : j \\in \\mathbb{Z} \\})$ with the evident difference field structure.<\/p>\n Proof:\u00a0 For the forward direction, let $L \\succeq K$ be an $|E|^+$-strongly homogeneous elementary extension of $K$.\u00a0 By strong homogeneity, if\u00a0 $\\operatorname{tp}(a\/E) = \\operatorname{tp}(b\/E)$, then there is an automorphism $\\tau:L \\to L$ fixing $E$ and taking $a$ to $b$.\u00a0 As $\\tau$ is a map of difference fields, it takes $\\sigma^j(a)$ to $\\sigma^j(b)$ for all $j \\in \\mathbb{Z}$.\u00a0 Thus, $\\tau$ maps $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}$ to $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}$, and thus anything algebraic over $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}$ to something algebraic over $E \\langle b \\rangle_{\\sigma,\\sigma^{-1}}$. That is, the restriction of $\\tau$ to $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ is an isomorphism of difference fields between $E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ and $E \\langle b \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ taking $a$ to $b$.<\/p>\n In the other direction, fix some isomorphism of difference fields $\\tau:E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg} \\to E \\langle b \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$ fixing $E$ and taking $a$ to $b$.\u00a0 Let $B := E \\langle a \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$.\u00a0 Let $L_1 := L_2 := K$.\u00a0 Embed $B$ into $K_1$ via the usual inclusion and into $L_2$ via $\\tau$.\u00a0 Then by our main proposition, $L_1 \\equiv_B L_2$.\u00a0 \u00a0Thus, for any formula $\\phi(x) \\in \\mathcal{L}_E(x)$ where $x$ is a free variable of length equal to that of $a$, we have $K \\models \\phi(a) \\Longleftrightarrow L_1 \\models \\phi(a) \\Longleftrightarrow L_2 \\models \\phi(a) \\Longleftrightarrow K \\models \\phi(\\tau(a)) \\Longleftrightarrow K \\models \\phi(b)$.\u00a0 That is, $\\operatorname{tp}(a\/E) = \\operatorname{tp}(b\/E)$. $\\Box$<\/p>\n Corollary:\u00a0<\/strong> If $K \\models \\text{ACFA}$, $k_0$ is the prime field in $K$, and $A \\subseteq K$ is any subset, then $\\operatorname{acl}(A) = k_0 \\langle A \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$.<\/p>\n Proof:\u00a0<\/strong> Set $E := k_0 \\langle A \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$.\u00a0 \u00a0 Clearly, every element of $E$ is algebraic over $A$.\u00a0 \u00a0\u00a0Suppose that $b \\in K \\smallsetminus E$.\u00a0 Let $L := E \\langle b \\rangle_{\\sigma,\\sigma^{-1}}^\\text{alg}$.\u00a0 Because $E$ is algebraically closed, for every $n$, the algebra $L^{\\otimes n} = L \\otimes_E L\u00a0 \\otimes_E\u00a0 \\cdots \\otimes_E L$ is an integral domain with an endomorphism $\\sigma$ defined on tensors by $a_1 \\otimes \\cdots \\otimes a_n \\mapsto \\sigma(a_1) \\otimes \\cdots \\otimes \\sigma(a_n)$.\u00a0 Hence, it may be embedded into a difference field, and then into a model $(M,\\sigma)$ of ACFA.\u00a0 The images of $b \\otimes 1 \\otimes \\cdots \\otimes 1$, $1 \\otimes b \\otimes 1 \\cdots \\otimes 1$, $\\ldots$, $1 \\otimes 1 \\otimes \\cdots \\otimes b$ are distinct but the isomorphism between $L$ and its image where $x$ is mapped to the tensor $1 \\otimes \\cdots \\otimes 1 \\otimes x \\otimes 1 \\otimes \\cdots \\otimes 1$ where $x$ appears as the $m^\\text{th}$ component shows by our last proposition that all of those elements have the same type over $E$ as $b$ itself.\u00a0 Thus, $b \\notin \\operatorname{acl}(E)$; showing that $E$ is model theoretically algebraically closed.\u00a0 $\\Box$<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" Our proof of model completeness for ACFA yields\u00a0 strong quantifier simplification results. Let us start with a relative completeness result.\u00a0 \u00a0If a theory $T$ is model complete, then whenever $E \\models T$ and $E \\subseteq M_i \\models T$ for $i … Continue reading