Proposition 7

— Let $\Omega$ be an algebraic extension of a field $K$. a) If $\Omega$ is algebraically closed, then every algebraic extension of $K$ is isomorphic to a subextension of $\mathbf{R}$. b) Conversely suppose that every algebraic extension of finite degree of $K$ is isomorphic to a subextension of $\Omega$; then $\Omega$ is algebraically closed.

Assertion a) follows from Th. 1 (V, p. 20). Now assume the hypotheses of b) and consider a non-constant polynomial $f \in K[X]$. Let E be a splitting field of $f$ (V, p. 21, Prop. 4); since E is algebraic of finite degree over $K$ (V, p. 18, Th. 2), we may suppose that E is a subextension of $\Omega$. Then the polynomial $f$ is a product of polynomials of degree 1 in $\Omega[X]$ and Prop. 6 shows that $\Omega$ is algebraically closed.

We can now prove the existence and uniqueness (up to isomorphism) of the algebraic closure of a field.