A note on geometric theories of fields

Otherwise, $K$ is an infinite field of characteristic $p$. Take $a\in K\setminus K^{p}$. Then the map

is a definable injection $f:K\times K\to K$. But $\dim(K^{2})=2>\dim(K)=1$, as $K$ is infinite. ∎

If $a\in K\setminus S$, then the function $f(x,y)=ax+y$ is injective on $X^{2}$. ∎

If $X$ is infinite, then $\dim(X)>0$, so $\dim(X^{2})\geq 2>\dim(K)=1$ and there are no definable injections $f:X^{2}\to K$. Therefore case (2) of Lemma 2.3 cannot hold, so (1) holds. Conversely, if $X$ is finite, then case (1) of Lemma 2.3 cannot hold, as the set $S$ would be finite. ∎

It suffices to eliminate $\exists^{\infty}$. If models of $T$ are finite, then $\exists^{\infty}$ is trivially eliminated. If the models of $T$ are infinite, then Lemma 2.4 gives a first-order criterion for telling whether a definable set $X\subseteq K$ is infinite. ∎

If $K$ is very slim, then $K$ is pregeometric, hence geometric by Theorem 2.5 and perfect by Lemma 2.2 (or [JK10, Proposition 4.1]). ∎

1. (1)

   Clearly $(AF)^{\mathrm{alg}}\cap K\subseteq\operatorname{acl}(AF)$. Conversely if $b\in\operatorname{acl}(AF)$, then $b$ is in a finite set $\varphi({\bar{a}},K)$ for some tuple ${\bar{a}}$ from $AF$. By algebraic boundedness, $\varphi({\bar{a}},K)$ is contained in a finite zero set of some polynomial $P({\bar{a}},y)$, where $P({\bar{x}},y)\in F[{\bar{x}},y]$. Therefore $b$ is field-theoretically algebraic over $AF$.

2. (2)

   For a fixed formula $\varphi({\bar{x}},y)$, let $P_{1},\ldots,P_{m}\in F[{\bar{x}},y]$ be polynomials as in Definition 2.10. Then the cardinality of finite sets of the form $\varphi({\bar{a}},K)$ is bounded by the maximum of the degrees of the $P_{i}$’s.

3. (3)

   By (2), the theory of $K$ has uniform finiteness, and so $\exists^{\infty}$ is uniformly eliminated across elementary extension of $K$. It follows that for fixed $\varphi,P_{1},\ldots,P_{m}$, the following condition is perserved in elementary extensions:

   For any ${\bar{a}}$, if $\varphi({\bar{a}},y)$ defines a finite set, then there is $i\in\{1,\ldots,m\}$ such that $P_{i}({\bar{a}},y)$ has finitely many zeros and $\varphi({\bar{a}},y)\rightarrow P_{i}({\bar{a}},y)=0$.

   Therefore algebraic boundedness transfers from $K$ to any elementary extension $K^{*}$. ∎

(1)$\implies$(2). Field-theoretic algebraic closure satisfies the exchange property. Therefore $\operatorname{acl}(-)$ satisfies exchange (after naming the elements of $F$ as parameters). By Theorem 2.5, elementary extensions of $K$ eliminate $\exists^{\infty}$.

(2)$\implies$(3). Suppose (2) holds but $K$ fails to be algebraically bounded over $F$, witnessed by some formula $\varphi({\bar{x}},y)$. Using elimination of $\exists^{\infty}$, we may assume that there is $n\in\mathbb{N}$ such that $|\varphi({\bar{a}},K)|\leq n$ for every ${\bar{a}}$. For any finite set of polynomials $P_{1},\ldots,P_{m}\in F[{\bar{x}},y]$, there is ${\bar{a}}\in K$ such that $\varphi({\bar{a}},K)$ is finite, but is not contained in the zero set of $P_{i}({\bar{a}},y)$ unless $P_{i}({\bar{a}},y)\equiv 0$. By compactness, there is an elementary extension $K^{*}\succeq K$ and a tuple ${\bar{a}}\in K^{*}$ such that $\varphi({\bar{a}},K^{*})$ is finite, but is not contained in the zero set of $P({\bar{a}},y)$ for any $P\in F[{\bar{x}},y]$ except those with $P({\bar{a}},y)\equiv 0$. Then $\varphi({\bar{a}},K^{*})$ contains a point not in $F({\bar{a}})^{\mathrm{alg}}$, contradicting (2).

(3)$\implies$(1). Lemma 2.11 shows that if $K^{*}$ is an elementary extension of $K$, then

• •

  $K^{*}$ is algebraically bounded over $F$.

• •

  Field-theoretic and model-theoretic algebraic agree over $F$.

Therefore (1) holds. ∎

We use the criterion of Lemma 2.12(1) to show that $K$ is algebraically bounded over $F$. Embed $K$ into a monster model $\mathbb{K}$. Suppose $b,{\bar{c}}\in\mathbb{K}$ and $b\in\operatorname{acl}(F\bar{c})$. We must show $b\in F({\bar{c}})^{\mathrm{alg}}$. By Remark 2.15, $\mathrm{Th}(K)$ is geometric, because it is geometric after naming parameters from $F$. Replacing ${\bar{c}}$ by a basis of ${\bar{c}}$ in the $\operatorname{acl}$-pregeometry, we may assume that the tuple ${\bar{c}}$ is field- theoretically algebraically independent over $F$. Now suppose for the sake of contradiction that $b\notin F({\bar{c}})^{\mathrm{alg}}$. Then $({\bar{c}},b)$ is also field-theoretically algebraically independent over $F$.

As $b\in\operatorname{acl}(F{\bar{c}})=\operatorname{acl}({\bar{c}})$, there is a 0-definable set $D\subseteq\mathbb{K}^{n}$ with $(b,{\bar{c}})\in D$ and $D_{{\bar{c}}^{\prime}}$ finite for each ${\bar{c}}^{\prime}$. By algebraic boundedness over $K$, there are finitely many non-zero polynomials $P_{1},\ldots,P_{i}\in K[x,{\bar{y}}]$ such that $D$ is contained in the union of the zero-sets of the $P_{i}$.

Let $M=\mathbb{K}^{\mathrm{alg}}$ and let $V$ be the Zariski closure of $D$ in $M^{n+1}$. The polynomials $P_{i}$ show that $V\subsetneq M^{n+1}$, and so $\dim(V)<n+1$. By elimination of imaginaries in ACF, there is a finite tuple $e$ in $M$ which codes $V$. Recall that $\mathbb{K}$ is perfect by Lemma 2.2. If $\sigma\in\operatorname{Aut}(M/\mathbb{K})=\operatorname{Gal}(\mathbb{K})$, then $\sigma(D)=D$, $\sigma(V)=V$, and $\sigma(e)=e$. As the tuple $e$ is fixed by $\operatorname{Gal}(M/\mathbb{K})$, it must be in the perfect field $\mathbb{K}$.

If $\sigma$ is any automorphism of $\mathbb{K}$, then $\sigma$ can be extended to an automorphism $\sigma^{\prime}$ of $M$. The fact that $D$ is 0-definable implies that $\sigma^{\prime}(D)=\sigma(D)=D$, which then implies $\sigma^{\prime}(V)=V$ and $\sigma(e)=\sigma^{\prime}(e)=e$. Thus $e$ is $\operatorname{Aut}(\mathbb{K})$-invariant, which implies that $e$ is in $F=\operatorname{dcl}(\varnothing)$. Therefore, in the structure $M$, the $e$-definable set $V$ is in fact $F$-definable. However, the tuple $(b,{\bar{c}})\in D\subseteq V$ is algebraically independent over $F$, so this implies $\dim(V)=n+1$, contradicting the earlier fact that $\dim(V)<n+1$. ∎

Clearly $|X|\leq|K|$. We must show $|X|\geq|K|$. Replacing $X$ with a projection onto one of the coordinate axes, we may assume $X\subseteq K^{1}$. By Lemma 2.4, $K=\{(y-y^{\prime})/(x^{\prime}-x):x,y,x^{\prime},y^{\prime}\in X,~x\neq x^{\prime}\}$. Therefore $|K|\leq|X|^{4}=|X|$. ∎

Note that (2) implies (1), by taking $D=g(Y)$. We prove (1). Embed $M$ into a monster model $\mathbb{M}\succeq M$. Take $e\in X$ with $\dim(e/M)=d$. Then $e\in\mathbb{M}^{\mathrm{eq}}$. Every imaginary is definable from a real tuple, so there is a real tuple ${\bar{a}}\in\mathbb{M}^{m}$ with $e\in\operatorname{acl}^{\mathrm{eq}}(M{\bar{a}})$. Replacing ${\bar{a}}$ with a subtuple, we may assume that ${\bar{a}}$ is independent over $M$. By [Gag05, Lemma 3.1], $\operatorname{acl}(-)$ continues to satisfy exchange after naming the parameter $e$. Therefore, we can meaningfully talk about real tuples being independent over $eM$. Write ${\bar{a}}$ as ${\bar{b}}{\bar{c}}$ where ${\bar{b}}$ is a maximal subtuple that is independent over $eM$. Then ${\bar{c}}\in\operatorname{acl}({\bar{b}}eM)$. At the same time, $e\in\operatorname{acl}({\bar{b}}{\bar{c}}M)$, so $e$ is interalgebraic with ${\bar{c}}$ over ${\bar{b}}M$.

Meanwhile, $\dim({\bar{b}}/eM)=\dim({\bar{b}}/M)=|{\bar{b}}|$, because ${\bar{b}}$ is an independent tuple over $eM$. Then ${\bar{b}}$ and $e$ are independent from each other over $M$, implying $\dim(e/{\bar{b}}M)=\dim(e/M)=d$ by symmetry. As $e$ and ${\bar{c}}$ are interalgebraic over ${\bar{b}}M$, we have $\dim({\bar{c}}/{\bar{b}}M)=d$. But ${\bar{b}}{\bar{c}}$ is an independent tuple over $M$, so $\dim({\bar{c}}/{\bar{b}}M)$ is the length of ${\bar{c}}$. Thus ${\bar{c}}\in M^{d}$.

As $e$ and ${\bar{c}}$ are interalgebraic over ${\bar{b}}M$, there is a ${\bar{b}}M$-interpretable set $Y_{0}\subseteq X\times\mathbb{M}^{d}$ such that $(e,{\bar{c}})\in Y_{0}$, and the projections $f_{0}:Y_{0}\to X$ and $g_{0}:Y_{0}\to\mathbb{M}^{d}$ have finite fibers. By saturation, there is a uniform bound $N$ on the fiber size. The image $g_{0}(Y_{0})$ is ${\bar{b}}M$-definable and contains the point ${\bar{c}}$ with $\dim({\bar{c}}/{\bar{b}}M)=d$. Therefore $g_{0}(Y_{0})$ has dimension $d$.

Now we have the desired configuration $(Y_{0},f_{0},g_{0})$, but defined over the parameter ${\bar{b}}$ outside $M$. Because $M\preceq\mathbb{M}$ and dimension is definable in families [Gag05, Fact 2.4], we can replace the parameter ${\bar{b}}$ with something in $M$, getting a $M$-definable configuration $(Y,f,g)$ in which the fibers of $f$ and $g$ are still bounded in size by $N$. ∎

Let $T$ be the elementary diagram of $M$. Let $\mathcal{L}$ be the language of $T$, and let $\mathcal{L}(Q)$ be the language obtained by adding a new quantifier $(Qx)$. Let $\psi\mapsto\psi^{*}$ be the map from $\mathcal{L}(Q)$-formulas to $\mathcal{L}$-formulas interpreting $(Qx)\varphi(x,{\bar{y}})$ as “the set of $x$ such that $\varphi(x,{\bar{y}})$ holds is large.” More precisely,

• •

  $\varphi^{*}=\varphi$ if $\varphi$ doesn’t involve the quantifier $Q$.

• •

  $(\varphi\wedge\psi)^{*}=\varphi^{*}\wedge\psi^{*}$, and similarly for the other logical operators including $\exists$ and $\forall$.

• •

  $((Qx)\varphi(x,{\bar{y}}))^{*}$ is the formula $\psi({\bar{y}})$ such that in models $N\models T$,

  $$N\models\psi({\bar{b}})\iff\left(\varphi^{*}(N,{\bar{b}})\text{ is large}\right).$$

Let $T^{\prime}$ be the set of $\mathcal{L}(Q)$-sentences $\varphi$ such that $T\vdash\varphi^{*}$. It is straightforward to verify that $T^{\prime}$ is closed under the rules of inference on pages 6–7 of [Kei70]. For example, the “axioms of $\mathcal{L}(Q)$” [Kei70, p. 6] correspond to the axioms in Definition 3.3. By the completeness theorem for $\mathcal{L}(Q)$ [Kei70, Section 2], there is an $\mathcal{L}$-structure $N$ which satisfies the sentences $T^{\prime}$, when $(Qx)$ is interpreted as “there are uncountably many $x$ such that…” Ignoring the sentences involving $Q$, we see that $N\models T$, and so $N\succeq M$. Finally, suppose $X=\varphi(N,{\bar{b}})$ is definable in $N$. Let $\psi({\bar{y}})$ be the $\mathcal{L}$-formula such that $\psi({\bar{b}})$ holds iff $\varphi(N,{\bar{b}})$ is large. Then $T^{\prime}$ contains the sentence

because its image under $(-)^{*}$ is the tautology

Therefore

Take a model $M\models T$. There is a definable notion of largeness on $M^{\mathrm{eq}}$ in which $X$ is large iff $\dim(X)>0$. The requirements of Definition 3.3 hold by properties of dimension in $M^{\mathrm{eq}}$ [Joh22, Propositions 2.8, 2.9, 2.12]. Applying Fact 3.4, we get an elementary extension $N^{\mathrm{eq}}\succeq M^{\mathrm{eq}}$ such that if $X$ is definable in $N^{\mathrm{eq}}$, then $X$ is uncountable iff $\dim(X)>0$. ∎

1. (1)

   Suppose $X$ has positive dimension. Lemma 3.2 shows that $|X|\geq|D|$ for some infinite definable set $D$. Then $|D|\geq|X|\geq|K|$ by Proposition 3.1. Finally, $|K|\geq|X|$ is clear.

2. (2)

   Lemma 3.5 gives an uncountable model $K$ in which every zero-dimensional interpretable set is countable. By downward Löwenheim-Skolem, we can replace $K$ with an elementary substructure of cardinality $\aleph_{1}$. ∎

By Lemma 2.2, $K$ is perfect. Hence any finite extension of $K$ is a simple extension. Let $X$ be the set of irreducible monic polynomials of degree $n$. We can regard $X$ as a definable subset of $K^{n}$ by identifying a polynomial $P(x)=x^{n}+c_{n-1}x^{n-1}+\cdots+c_{0}$ with the $n$-tuple $(c_{0},c_{1},\ldots,c_{n-1})$. Let $Y$ be the interpretable set of degree $n$ finite extensions. Let $f:X\to Y$ be the map sending $P(x)$ to the extension $K[x]/(P(x))$.

Note that $\dim(X)\leq\dim(K^{n})=n$. By the definition of dimension for interpretable sets, it suffices to show that each fiber of $f$ has dimension at least $n$. Fix some $b\in Y$ corresponding to a degree $n$ extension $L/K$. We claim $f^{-1}(b)$ has dimension $n$. By identifying $L$ with $K^{n}$, we can regard $L$ as a definable set with $\dim(L)=n$. Because $K$ is perfect, there are only finitely many fields between $K$ and $L$. Let $U$ be the union of the intermediate fields. Then $U$ has lower dimension than $n$, so $\dim(L\setminus U)=n$. The elements of $L\setminus U$ are generators of $L$. The fiber $f^{-1}(b)$ is the set of minimal polynomials of elements of $L\setminus U$. Let $\rho:(L\setminus U)\to f^{-1}(b)$ be the map sending $a\in L\setminus U$ to its minimal polynomial. Then $\rho$ is finite-to-one, so $\dim(f^{-1}(b))\geq\dim(L\setminus U)=n$. ∎

Zero-dimensional definable sets are finite. If elimination of imaginaries holds, then zero-dimensional interpretable sets are finite. ∎