A field is algebraically closed exactly when every nonconstant polynomial splits, equivalently when it has no nontrivial finite extension

Assume $F$ is algebraically closed, and let $f\in F[x]$ be nonconstant. By [L1] it has a root $a\in F$, so [L3] gives $f=(x-a)q$ for some $q\in F[x]$. Repeating the same argument on $q$ while it remains nonconstant writes $f$ as a product of linear factors, so $f$ splits over $F$ in the sense of [L2].

Assume every nonconstant polynomial in $F[x]$ splits over $F$, and let $E\mathrm{/}F$ be finite. For any $\alpha \in E$, fact [L6] makes $\alpha$ algebraic over $F$. If $\alpha \mathrm{\notin }F$, let $m_{\alpha }$ be its minimal polynomial over $F$. Then [L7] makes $m_{\alpha }$ a nonconstant irreducible polynomial with $m_{\alpha }(\alpha )=0$. By the hypothesis, $m_{\alpha }$ splits over $F$ and therefore has a root $a\in F$. Fact [L3] gives $x-a$ as a linear factor of $m_{\alpha }$, contradicting irreducibility. Hence every $\alpha \in E$ already lies in $F$, so $E=F$.

Assume $F$ has no nontrivial finite extension, and let $f\in F[x]$ be nonconstant. By [L5] there is an extension $K\mathrm{/}F$ and an element $\alpha \in K$ with $f(\alpha )=0$. Then $\alpha$ is algebraic over $F$, so [L4] makes $F(\alpha )\mathrm{/}F$ finite. By the hypothesis this finite extension must be trivial, hence $\alpha \in F$. Therefore every nonconstant polynomial over $F$ has a root in $F$, and [L1] says that $F$ is algebraically closed.

Steps 1.1, 1.2, and 1.3 prove $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$, so the three conditions are equivalent.