Algebraically closed fields - Model Theory

Definition 9.1 (ACF). $ACF$ is the $L_{rings}$ theory axiomatising algebraically closed fields. It consists of:

field axioms
for every $d \geq 1$ , we add an axiom:
$\forall v_{0}, \dots, v_{d}, \exists x (v_{d} \neq 0 ⟹ v_{0} + v_{1} x + \dots + v_{d} x^{d} = 0) .$

Definition 9.2 (ACF with characteristic). For $n \geq 1$ , let $χ_{n}$ be the sentence

\underset{n times}{\underset{⏟}{1 + \dots + 1}} = 0 .

Set

\begin{array}{l} {ACF}_{0} & = ACF \cup {\neg χ_{n} : n \in ℕ} \\ {ACF}_{p} & = ACF \cup {χ_{p}} & (for p a prime) \end{array}

Proof. The Transcendence degree of $k ⊨ ACF$ (and algebraically closed field) is the cardinality of the largest algebraically independent subset of $k$ .

For example,

\begin{array}{l} {trdeg}_{ℚ} (\bar{ℚ}) & = 0 \\ {trdeg}_{ℚ} (\bar{ℚ (π)}) & = 1 \\ {trdeg}_{ℚ} (ℂ) & = 2^{ℵ_{0}} \\ {trdeg}_{ℚ} (\bar{ℚ {(x_{i})}_{i < κ}}) & = κ \end{array}

From algebra we know:

(1) If $k, k^{'} ⊨ ACF$ then $k ≅ k^{'}$ if and only if
- $trdeg (k) = trdeg (k^{'})$
- $char (k) = char (k^{'})$
- $| k | = | k^{'} |$
(2) If $k ⊨ ACF$ , $λ = {trdeg}_{ℚ} (k)$ , then $| κ | = ℵ_{0} + λ$ . Thus if $k, k^{'} ⊨ {ACF}_{0}$ or ${ACF}_{p}$ are uncountable and $| k | = | k^{'} |$ then ${trdeg}_{ℚ} (k) = {trdeg}_{ℚ} (k^{'})$ so $k ≅ k^{'}$ . □

Proof. Los-Vaught test. □

Definition 9.5 (Polynomial map). Let $k$ be a field. We say a function $ϕ : K^{m} \to K^{m}$ is a polynomial map if

Φ (\bar{x}) = (p_{1} (x_{1}, \dots, x_{m}), \dots, p_{n} (x_{1}, \dots, x_{m}))

where each $p_{1}, \dots, p_{n}$ is a polynomial.

Theorem 9.6 (Ax-Grothendieck). Assuming that:

$k$ an algebraically closed field
$Φ : K^{n} \to K^{n}$ an injective polynomial map

Then

Φ

is surjective.

Proof. First suppose $K = \bar{𝔽_{p}}$ for some prime $p$ . $\bar{𝔽_{p}} = ⋃_{k} 𝔽_{p^{k}}$ . Fix an $m$ such that the coefficients of $Φ$ all lie in $𝔽_{p^{m}}$ . Note: $\bar{𝔽_{p}} = ⋃_{k} 𝔽_{p^{m k}}$ .

For any $k \geq 1$ , $Φ$ induces an injective polynomial map $𝔽_{p^{k m}} \to 𝔽_{p^{k m}}$ which has to be surjective (as finite field). Hence

\begin{array}{l} Φ ({\bar{𝔽_{p}}}^{n}) & = Φ (⋃_{k} 𝔽_{p^{m k}}^{n}) \\ = ⋃_{k} Φ (𝔽_{p^{m k}}^{n}) \\ = ⋃_{k} 𝔽_{p^{m k}}^{n} \\ = {\bar{𝔽_{p}}}^{n} \end{array}

So $Φ$ is surjective.

Given $n, d \geq 1$ , let $ψ_{n, d}$ be the $L$ -sentence expressing “every injective polynomial map with $n$ -coordinates, each of which is a polynomial in $n$ variables with degree at most $d$ , is surjective”.

Exercise: Show that this is a first order $L_{rings}$ sentence.

Now $\bar{𝔽_{p}} ⊨ ψ_{n, d}$ for all $n, d \geq 1$ and $p$ prime.

${ACF}_{p}$ is complete, so ${ACF}_{p} ⊨ ψ_{n, d}$ . Now consider ${ACF}_{0}$ . Suppose for contradiction that ${ACF}_{0} ⁄ ⊨ ψ_{n, d}$ for some $n, d$ , so ${ACF}_{0} ⊨ \neg ψ_{n, d}$ .

By compactness, there exists $Σ \subseteq {ACF}_{0}$ finite such that $Σ ⊨ \neg ψ_{n, d}$ .

In particular, $Σ \subseteq ACF \cup {\neg χ_{0}, \dots, \neg χ_{m}}$ for some $m$ . Choose a prime $p$ such that $p > m$ and ${ACF}_{p} ⊢ Σ$ .

So must have ${ACF}_{p} ⊨ \neg ψ_{n, d}$ , contradiction. □

Theorem 9.7 (Lipschiptz principal). Assuming that:

$ϕ$ an $L_{rings}$ sentence

Then the following are equivalent:

(1) ${ACF}_{0} ⊨ ϕ$ , i.e. $ϕ$ in every $k ⊨ {ACF}_{0}$
(2) ${ACF}_{0} \cup {ϕ}$ is consistent
(3) there exists some $n > 0$ such that ${ACF}_{p} ⊨ ϕ$ for any $p > n$
(4) for all $n > 0$ , there exists some $p > n$ such that ${ACF}_{p} \cup {ϕ}$ is consistent

9 Algebraically closed fields