Proof. sketch [1601A]

1 $\iff$ 2 $\iff$ 3 is obvious.

3 $\implies$ 4: Take $\alpha \in K$ and consider $m_{\alpha, \Omega}(x)$. It must be $m_{\alpha, \Omega}(x) = x - \alpha \in \Omega[x]$ since it is irreducible and monic, hence $\alpha \in \Omega$. So we conclude that $K = \Omega$.

4 $\implies$ 5: Obvious because of Proposition 14010.

5 $\implies$ 1: Take any $f(x) \in \Omega[x]$ and let $K$ be its splitting field. Then by Proposition 1507, $[K : \Omega]$ is at most $n!$ where $n = \deg f(x)$, hence it is finite. So $K = \Omega$. Then we can conclude that every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.