Lefschetz principle

Remarks

Clauses 1, 2 and 3 are equivalent because the theory $\mathrm{ACF}_0$ of algebraically closed fields of characteristic $0$ is complete. Suppose $\varphi$ holds in $\mathrm{ACF}_0$. By the compactness theorem, $\varphi$ follows from $\mathrm{ACF}$ and finitely many of the axioms $1 + \cdots + 1 \ne 0$. So it holds in characteristic $p$ for all large $p$, which is clause 4. Clause 4 gives 5, since there are infinitely many primes. If $\varphi$ fails in $\mathrm{ACF}_0$, then $\neg \varphi$ holds there, so $\neg \varphi$ holds in all large characteristics and 5 fails.

Example: fix $n$ and $d$. The statement "every injective polynomial map $K^n \to K^n$ whose components have degree at most $d$ is surjective" is a first-order sentence about the field $K$. It holds in the algebraic closure of each finite field $\F_p$, so by clause 5 it holds in $\C$. This is the model-theoretic proof of the Ax–Grothendieck theorem.

The principle holds only for first-order sentences. The statement "every element is algebraic over the prime field" holds in the algebraic closure of each $\F_p$ but fails in $\C$. It is not expressible by a single first-order sentence.

Going deeper

For a student

The principle lets one prove algebraic facts about $\C$ by working over finite fields, where counting arguments are available, or the other way around. It comes from the completeness of $\mathrm{ACF}_p$ for each $p$, which itself follows from quantifier elimination or from uncountable categoricity and the Łoś–Vaught test. Take the algebraic closures of the fields $\F_p$ and a non-principal ultrafilter on the primes. Their ultraproduct is an algebraically closed field of characteristic $0$, which gives another view of clause 4.