Proposition 6

— *Let $\Omega$ be an extension of a field K. For $\Omega$ to be an algebraic closure of K it is necessary and sufficient that it should be algebraic and that each non-constant polynomial in $K[X]$ should split in $\Omega[X]$ into a product of factors of degree 1.

The condition is necessary by (AC). Conversely, suppose that $\Omega$ is algebraic over $K$ and every non-constant polynomial over $K[X]$ is a product in $\Omega[X]$ of factors of degree 1. Let $\Omega'$ be an algebraic extension of $\Omega$ and let $x \in \Omega'$. Since $x$ is algebraic over $\Omega$ and $\Omega$ is algebraic over $K$, $x$ is algebraic over $K$ (V, p. 19, Prop. 3). Let $f$ be the minimal polynomial of $x$ over $K$. By hypothesis the polynomial $f \in K[X]$ splits in $\Omega[X]$ into a product of factors of degree 1, whence $x \in \Omega$. Thus we have $\Omega' = \Omega$ and $\Omega$ is algebraically closed because it satisfies (AC).