Lefschetz Principle (First-Order)

Theorem
Let $\phi$ be a sentence in the language $\mathcal L_r = \left\{{0, 1, +, -, \cdot}\right\}$ for rings, where $0, 1$ are constants and $+, -, \cdot$ are binary functions.


 * $(1): \quad \phi$ is true in every algebraically closed field of characteristic $0$.
 * $(2): \quad \phi$ is true in some algebraically closed field of characteristic $0$.
 * $(3): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for arbitrarily large primes $p$.
 * $(4): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for sufficiently large primes $p$.

Note in particular that since $\C$ is an algebraically closed field of characteristic $0$, these are equivalent to $\phi$ being true in the field $\C$.

$(1)$ iff $(2)$
From Theory of Algebraically Closed Fields of Characteristic $p$ is Complete:

the theory $ACF_p$ of algebraically closed fields of characteristic $p$ is complete.

That is, all such fields satisfy the exact same $\mathcal L_r$ sentences.

$(2)$ implies $(4)$
Let $\phi$ be true in some such field.

Then:
 * $ACF_0 \models \phi$

By Gödel's Completeness Theorem and the finiteness of proofs, it follows that there is a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \phi$.

Such a $\Delta$ can only make finitely many assertions about the character of its models.

Hence, as long as $p$ is selected sufficiently large, an algebraically closed field of characteristic $p$ will satisfy $\phi$.

$(4)$ implies $(3)$
We have that all sufficiently large $p$ work.

Hence it follows that it is always possible to find arbitrarily large $p$ that work.

$(3)$ implies $(2)$
We prove this by contraposition.

Suppose there is no algebraically closed field of characteristic $0$ where $\phi$ is true.

Then $\phi$ is false in algebraically closed fields of characteristic $0$

Since $ACF_0$ is complete, this means that $ACF_0 \models \neg \phi$.

Similarly to the case of $(2)$ implies $(4)$, there must then be a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \neg \phi$.

But then, for all sufficiently large $p$, we have that $\phi$ is false in the algebraically closed fields of characteristic $p$.

Hence, it cannot be true for arbitrarily large $p$.

By Rule of Transposition, $\phi$ is true in algebraically closed fields for arbitrarily large primes $p$ implies there exists an algebraically closed field of characteristic $0$ where $\phi$ is true.