Theorem 1 (Zariski)

— *Let K be a field, L an extension of K, $K_1$ the relative separable algebraic closure of K in L (V, p. 44), $K_2$ the relative algebraic closure of K in L (V, p. 19) and $(X_i)_{i \in I}$ a family of indeterminates. Then $K_1(X_i)_{i \in I}$ is the relative separable algebraic closure of $K(X_i)_{i \in I}$ in $L(X_i)_{i \in I}$, and $K_2(X_i)_{i \in I}$ is the relative algebraic closure of $K(X_i)_{i \in I}$ in $L(X_i)_{i \in I}$.

A) Suppose that E is a field and F an extension field of E and that every element of F which is separable algebraic over E belongs to E. Let $u$ be an element of $F(X)$ which is separable algebraic over $E(X)$; we shall show that $u$ belongs to $E(X)$. There exist in $F(X)$ two relatively prime polynomials P and Q such that $u = P/Q$ and we may take Q to be monic. Let S be the finite subset of F consisting of the coefficients of P and Q, $F_0 = E(S)$ and let A be an E-derivation of $F_0$ into $F_0$. Let $D$ be the derivation of $F_0(X)$ into itself which agrees with $A$ on $F_0$ and maps $X$ to $0$ (V, p. 128, Prop. 3).

Since $u \in F_0(X)$ is separable algebraic over $E(X)$ and $D$ is zero on $E(X)$, we have $D(u) = 0$ (V, p.129, Prop. 4), whence $D(P) . Q = P . D(Q)$. Since $P$ and $Q$ are relatively prime, we conclude that $Q$ divides $D(Q)$ (IV, p. 13, Cor. 4). Now $Q$ may be written in the form

with $a_1, \ldots, a_n$ in $F_0$; since $D(x) = 0$, we thus have whence $\deg D(Q) < \deg Q$. Since $Q$ divides $D(Q)$, this is possible only if $D(Q) = 0$. But then $D(P) = 0$ because $D(P) . Q = P . D(Q)$. Now (1) and a similar formula for $P$ show that $A$ annihilates the set $S$ of coefficients of $P$ and $Q$, whence $A = 0$, because $F_0 = E(S)$. By V, p. 135, Cor. 2, the finitely generated extension $F_0$ of $E$ is thus separable algebraic; by the hypotheses on $E$ and $F$ we have $F_0 = E$, so finally $u \in E(X)$.

B) Suppose now that $E$ is relatively algebraically closed in the extension field $F$ and denote by $p$ the characteristic exponent of $E$. Let $u$ be an element of $F(X)$ which is algebraic over $E(X)$. There exists an integer $f \geq 0$ such that $v = u^{p^f}$ is separable algebraic over $E(X)$ (V, p. 44, Prop. 13). By A) we thus have $v \in E(X)$. There exists a unique representation of $u$ in the form $P/Q$ with relatively prime polynomials $P$ and $Q$ in $F[X]$ and $Q$ monic; we have a similar decomposition $v = P_1/Q_1$ with $P_1$ and $Q_1$ relatively prime in $E[X]$ and $Q_1$ monic. It follows that $P_1/Q_1 = P^{p^f}/Q^{p^f}$; the polynomials $P^{p^f}$ and $Q^{p^f}$ are relatively prime in $F[X]$ (IV, p. 13, Cor. 6), like $P_1$ and $Q_1$, and $Q^{p^f}$ is monic. It follows that $P^{p^f} = P_1 \in E[X]$ and $Q^{p^f} = Q_1 \in E[X]$. Therefore the coefficients of $P$ and $Q$ are $p$-radical over $E$, and hence belong to $E$ because $E$ is relatively algebraically closed in $F$. We thus have $P \in E[X]$, $Q \in E[X]$ and finally $u \in E(X)$. This proves $E(X)$ to be relatively algebraically closed in $F(X)$.

C) We use the notation of Theorem 1. Since $K_1$ is a separable algebraic extension of $K$, the extension $K_1(X_i)_{i \in I}$ of $K(X_i)_{i \in I}$, is algebraic and separable (V, p. 39, Prop. 6). Moreover, every element of $L$ which is algebraic and separable over $K$, belongs to $K$, (V, p. 44, Prop. 13, a)). Let $J$ be a finite subset of $I$; by an immediate induction on the cardinal of $J$ we deduce from A) that every element of $L(X_i)_{i \in I}$, which is algebraic and separable over $K(X_i)_{i \in I}$, belongs to $K_1(X_i)_{i \in J}$. Finally let $u$ be an element of $L(X_i)_{i \in I}$, algebraic and separable over $K(X_i)_{i \in I}$; there exists a finite subset $J$ of $I$ such that $u$ belongs to $L(X_i)_{i \in I}$ and is algebraic and separable over $K(X_i)_{i \in J}$; by what has been said, $u$ belongs to $K_1(X_i)_{i \in J}$, and a fortiori to $K_1(X_i)_{i \in I}$.

We have thus deduced from A) that $K_1(X_i)_{i \in I}$ is the relative separable closure of $K(X_i)_{i \in I}$ in $L(X_i)_{i \in I}$; in the same way it follows from B) that $K_2(X_i)_{i \in I}$ is the relative algebraic closure of $K(X_i)_{i \in I}$ in $L(X_i)_{i \in I}$.