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Proof. If $F$ is algebraically closed, then every irreducible polynomial is linear. Namely, if there exists an irreducible polynomial of degree $> 1$, then this generates a nontrivial finite (hence algebraic) field extension, see Example 9.7.6. Thus (1) implies (2). If every irreducible polynomial is linear, then every irreducible polynomial has a root, whence every nonconstant polynomial has a root. Thus (2) implies (3).

Assume every nonconstant polynomial has a root. Let $P \in F[x]$ be nonconstant. If $P(\alpha ) = 0$ with $\alpha \in F$, then we see that $P = (x - \alpha )Q$ for some $Q \in F[x]$ (by division with remainder). Thus we can argue by induction on the degree that any nonconstant polynomial can be written as a product $c \prod (x - \alpha _ i)$.

Finally, suppose that every nonconstant polynomial over $F$ is a product of linear factors. Let $E/F$ be an algebraic extension. Then all the simple subextensions $F(\alpha )/F$ of $E$ are necessarily trivial (because the only irreducible polynomials are linear by assumption). Thus $E = F$. We see that (4) implies (1) and we are done. $\square$