[proofplan]
We pass from the parameter set $A$ to the subfield $k$ generated by $A$, whose cardinality is still at most $|A|$. [Quantifier elimination for algebraically closed fields](/theorems/4310) reduces every formula in one variable over $A$ to a Boolean combination of polynomial equations over $k$. Thus a complete $1$-type is determined by which polynomials in $k[X]$ vanish at a realization. In one variable, the possible prime vanishing ideals are either $(0)$, giving the unique transcendental type, or $(f)$ for an irreducible polynomial $f \in k[X]$, giving at most $|k|$ algebraic types.
[/proofplan]
[step:Replace the parameter set by the subfield it generates]
Let $\mathcal{L}_{\mathrm{ring}} := \{0,1,+,-,\cdot\}$ denote the language of rings. Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive natural numbers, and let $\aleph_0 := |\mathbb{N}|$ denote the cardinality of a countably infinite set. Let $k \subseteq K$ be the subfield generated by $A$. Since $A$ is infinite, every element of $k$ is obtained from finitely many elements of $A$ using finitely many applications of the field operations, so
\begin{align*}
|k| \leq |A| + \aleph_0 = |A|.
\end{align*}
The [polynomial ring](/page/Polynomial%20Ring) $k[X]$ has cardinality
\begin{align*}
|k[X]| = |k|.
\end{align*}
Indeed, $k[X]$ is the union over $n \in \mathbb{N}$ of $k^{n+1}$, the set of coefficient tuples of polynomials of degree at most $n$, and for infinite $k$ this countable union has cardinality $|k|$.
[guided]
Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive natural numbers, and let $\aleph_0 := |\mathbb{N}|$ denote the cardinality of a countably infinite set. Let $k$ be the smallest subfield of $K$ containing $A$. This is the right object because formulas with parameters from $A$ can involve field terms built from those parameters, and those terms evaluate inside the field generated by $A$.
We verify the cardinal estimate carefully. Each element of $k$ is represented by a rational expression in finitely many elements of $A$ with coefficients from the prime field of characteristic $p$. There are only $|A|$ many finite tuples from $A$, because $A$ is infinite, and only countably many formal rational expressions in finitely many variables over the prime field. Hence
\begin{align*}
|k| \leq |A| \cdot \aleph_0 = |A|.
\end{align*}
Next, $k[X]$ has cardinality $|k|$. For each $n \in \mathbb{N}$, polynomials of degree at most $n$ are encoded by coefficient tuples in $k^{n+1}$, so that set has size $|k|$. Taking the countable union over all $n$ gives
\begin{align*}
|k[X]| \leq \aleph_0 \cdot |k| = |k|.
\end{align*}
The reverse inequality follows because constant polynomials identify $k$ with a subset of $k[X]$. Therefore $|k[X]| = |k|$.
[/guided]
[/step]
[step:Use quantifier elimination to reduce types to polynomial vanishing data]
We use the standard quantifier elimination theorem for algebraically closed fields: every $\mathcal{L}_{\mathrm{ring}}(A)$-formula in one free variable is equivalent modulo $\operatorname{ACF}_p$ to a Boolean combination of equations
\begin{align*}
f(X) = 0,
\end{align*}
where $f \in k[X]$; citing a result not yet in the wiki: Quantifier Elimination for Algebraically Closed Fields.
Let $q \in S_1(A)$ be a complete $1$-type over $A$. Define its polynomial vanishing ideal
\begin{align*}
I_q := \{f \in k[X] : f(X) = 0 \in q\}.
\end{align*}
Because $q$ is complete and consistent with the field axioms, $I_q$ is a prime ideal of $k[X]$. By quantifier elimination, $q$ is determined by $I_q$: for each $f \in k[X]$, the type decides whether $f(X)=0$ or $f(X)\neq 0$, and every formula over $A$ is equivalent to a Boolean combination of such decisions.
[/step]
[step:Classify the possible polynomial vanishing ideals]
Since $k$ is a field, $k[X]$ is a [principal ideal domain](/page/Principal%20Ideal%20Domain). Therefore every prime ideal of $k[X]$ is either $(0)$ or $(f)$ for some irreducible polynomial $f \in k[X]$.
If $I_q = (0)$, then $q$ contains
\begin{align*}
f(X) \neq 0
\end{align*}
for every nonzero $f \in k[X]$. This is the unique type of an element transcendental over $k$.
If $I_q = (f)$ for an irreducible polynomial $f \in k[X]$, then $q$ contains $f(X)=0$ and, for every $g \in k[X]$,
\begin{align*}
g(X)=0 \in q \quad \Longleftrightarrow \quad f \mid g.
\end{align*}
Thus the algebraic types over $A$ are indexed by irreducible polynomials in $k[X]$, up to the harmless overcounting of assigning at most one type to each such polynomial.
[/step]
[step:Count the types and conclude stability]
There is one transcendental type and at most $|k[X]|$ algebraic types. Hence
\begin{align*}
|S_1(A)| \leq 1 + |k[X]| = |k| \leq |A|.
\end{align*}
This proves the asserted bound for every infinite $A \subseteq K$.
Finally, let $\kappa$ be an infinite cardinal with $\kappa \geq |\mathcal{L}_{\mathrm{ring}}|$, and let $A$ be any parameter set of size $\kappa$ in a model of $\operatorname{ACF}_p$. Applying the bound just proved gives
\begin{align*}
|S_1(A)| \leq \kappa.
\end{align*}
Therefore $\operatorname{ACF}_p$ is $\kappa$-stable for every such $\kappa$.
[/step]
