[proofplan]
We first show that elementary equivalence preserves characteristic by writing characteristic conditions as first-order ring sentences. In positive characteristic, the sentence asserting $p \cdot 1 = 0$ detects characteristic $p$ among fields; in characteristic $0$, the infinite family of sentences asserting $n \cdot 1 \ne 0$ for every $n \ge 1$ detects characteristic $0$. Conversely, if two algebraically closed fields have the same characteristic, then they are models of the same complete first-order theory $\mathrm{ACF}_p$ or $\mathrm{ACF}_0$, and completeness implies elementary equivalence.
[/proofplan]
[step:Express characteristic as first-order ring sentences]
For each integer $n \ge 1$, let $\sigma_n$ denote the $\mathcal{L}_{\mathrm{ring}}$-sentence
\begin{align*}
\underbrace{1 + 1 + \cdots + 1}_{n\text{ summands}} = 0.
\end{align*}
For a field $F$, the sentence $\sigma_n$ holds in $F$ exactly when $n \cdot 1_F = 0_F$.
If $\operatorname{char}(F) = p > 0$, then $p$ is prime and $F \models \sigma_p$. Moreover, for a prime number $q$, $F \models \sigma_q$ if and only if $p$ divides $q$, hence if and only if $q = p$.
If $\operatorname{char}(F) = 0$, then $F \models \neg \sigma_n$ for every integer $n \ge 1$.
[/step]
[step:Use elementary equivalence to identify the characteristic]
Assume $K \equiv L$. We prove $\operatorname{char}(K) = \operatorname{char}(L)$.
If $\operatorname{char}(K) = p > 0$, then $K \models \sigma_p$. Since $K \equiv L$, the fields satisfy the same first-order $\mathcal{L}_{\mathrm{ring}}$-sentences, so $L \models \sigma_p$. Therefore $p \cdot 1_L = 0_L$, and hence $\operatorname{char}(L)$ is a positive prime divisor of $p$. Since $p$ is prime, $\operatorname{char}(L)=p$.
If $\operatorname{char}(K) = 0$, then $K \models \neg \sigma_n$ for every integer $n \ge 1$. Elementary equivalence gives $L \models \neg \sigma_n$ for every integer $n \ge 1$, so no positive integer annihilates $1_L$. Hence $\operatorname{char}(L)=0$.
Thus $K$ and $L$ have the same characteristic.
[/step]
[step:Place fields of the same characteristic in the same complete theory]
Assume $\operatorname{char}(K) = \operatorname{char}(L)$. Let $c$ denote this common characteristic, where $c=0$ or $c=p$ for a prime number $p$.
If $c=p>0$, then both $K$ and $L$ are models of the first-order theory $\mathrm{ACF}_p$ of algebraically closed fields of characteristic $p$. If $c=0$, then both $K$ and $L$ are models of the first-order theory $\mathrm{ACF}_0$ of algebraically closed fields of characteristic $0$.
We use the completeness theorem for algebraically closed fields of fixed characteristic: for each prime $p$, the theory $\mathrm{ACF}_p$ is complete, and the theory $\mathrm{ACF}_0$ is complete. (citing a result not yet in the wiki: Completeness of Algebraically Closed Fields of Fixed Characteristic)
[/step]
[step:Apply completeness to obtain elementary equivalence]
Let $T$ be $\mathrm{ACF}_p$ if the common characteristic is a prime $p$, and let $T$ be $\mathrm{ACF}_0$ if the common characteristic is $0$. From the preceding step, $K \models T$ and $L \models T$.
Since $T$ is complete, for every $\mathcal{L}_{\mathrm{ring}}$-sentence $\varphi$, exactly one of $\varphi$ or $\neg \varphi$ belongs to $T$. Therefore, if $K \models \varphi$, then $T \not\models \neg \varphi$, so completeness forces $T \models \varphi$, and hence $L \models \varphi$. The same argument with $K$ and $L$ interchanged shows that $K$ and $L$ satisfy exactly the same first-order $\mathcal{L}_{\mathrm{ring}}$-sentences.
Thus $K \equiv L$, completing the proof of both directions.
[/step]
