Proposition 1

— For an extension $E$ of $K$ to be algebraic it is necessary and sufficient that every sub-K-algebra $A$ of $E$ should be a field.

$^1$ Cf. for example D. Hilbert, Gesammelte Abhandlungen, vol. 1, p. 1 (Berlin (Springer), 1932).

The condition is necessary : if $E$ is algebraic over $K$ and $x \neq 0$ is an element of a sub-K-algebra $A$ of $E$, then $x^{-1} \in K[x] \subset A$ by $V$, p. 16, Th. 1, d). Therefore $A$ is a field.

The condition is sufficient : if it is satisfied and $x$ is an element $\neq 0$ in $E$, then the ring $K[x]$ is a field, hence $x^{-1} \in K[x]$; in other words, there exists a polynomial $g \in K[X]$ such that $x^{-1} = g(x)$, or also $xg(x) - 1 = 0$; this shows $x$ to be algebraic over $K$, hence $E$ is an algebraic extension of $K$.