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 $L(X_i)_{i \in I}$.

On replacing $\Omega$ by an algebraic closure if necessary, we may suppose $\Omega$ algebraically closed. Let $B$ be a transcendence basis of $M$ over $K$, $N$ the relative algebraic closure of $K(B)$ in $\Omega$ and $N_r$ (resp. $N_s$) the set of elements of $N$ which are separable (resp. p-radical) over $K(B)$. We remark that $M$ is algebraic over $K(B)$, so that $N$ is the relative algebraic closure of $M$ in $\Omega$.

Fig. 1.

Let us define the following chain of homomorphisms:

We have $\alpha = \mathrm{Id}_L \otimes u$ where $u$ is the canonical injection of $M$ into $N$, hence $\alpha$ is injective. The mapping $\beta$ is the isomorphism of commutative groups which maps $x \otimes y$ to $(x \otimes 1) \otimes y$ (II, p. 278, Prop. 2) for $x \in L$ and $y \in N$. We have $\gamma = v \otimes \mathrm{Id}_N$, where $v$ is the $K$-algebra homomorphism of $L \otimes_K K(B)$ into $L(B)$ which maps $x \otimes y$ to $xy$, for $x \in L$ and $y \in K(B)$; since $L$ and $M$ are algebraically disjoint over $K$, Prop. 14 (V, p. 114) shows that $L$ and $K(B)$ are linearly disjoint over $K$, in other words, $v$ is injective, hence $y$ is injective. Since $N$ is a quasi-Galois extension of $K(B)$, there exists (V, p. 76, Prop. 13) a $K(B)$-algebra isomorphism $w$ of $N_s \otimes_{K(B)} N_r$ onto $N$ mapping $x \otimes y$ to

1 This hypothesis is sometimes expressed by saying that $L$ is a primary extension of $K$.

xy for $x \in \mathbf{N}$, and $y \in \mathbf{N}_r$; we denote by 6 the isomorphism $\mathrm{Id}_{L(B)} \otimes w \ ^1$. By Th. 1 (V, p. 137) and the hypothesis on the extension L of K, every element of $L(B)$ which is algebraic and separable over $K(B)$ belongs to $K(B)$; in particular, we have $L(B) \cap N_s = K(B)$. Since $N_r$ is a Galois extension of $K(B)$, Th. 5 (V, p. 71) shows that there exists a $K(B)$-algebra isomorphism $w'$ of $L(B) \otimes_{K(B)} N_s$ onto $L(B U N_r)$ mapping $x \otimes y$ to $xy$ for $x \in L(B)$ and $y \in N_r$; we denote by $\varepsilon$ the isomorphism $w' \otimes \mathrm{Id}_{N_r}$. Finally, $\zeta$ is the K-algebra homomorphism mapping $x \otimes y$ to $xy$ for $x \in L(B U N_r)$ and $y \in N_r$.

What has been said shows that $\eta = \varepsilon \delta \gamma \beta \alpha$ is an injective K-algebra homomorphism of $L \otimes_K M$ into $L(B U N_r) \otimes_{K(B), N_r}$. Further, every element of M is of the form $\sum_{i=1}^n a_i b_i$ with $a_i \in \mathbf{N}$, and $b_i \in \mathbf{N}_r$ for $1 \leq i \leq n$; hence we obtain $\varphi = \zeta \eta$.

The kernel $p$ of $\varphi$ is a prime ideal of $L \otimes_K M$, hence every nilpotent element of $L \otimes_K M$ belongs to $p$, by Prop. 2 (V, p. 118). Conversely let $a$ be an element of $p$; put $\eta(a) = \sum_{r=1}^s b_i \otimes c_i$ with $b_i \in L(B U N_r)$ and $c_i \in N_r$ for $1 \leq i \leq s$. Since $N_r$ is a p-radical extension of $K(B)$, there exists an integer $f \geq 0$ such that $c_i^{p^f}$ belongs to $K(B)$ for $1 \leq i \leq s$ (where $p$ is the characteristic exponent of K). But we have

and since $\eta$ is injective we finally have $a^{p^f} = 0$. We have thus shown $p$ to be the set of all nilpotent elements of $L \otimes_K M$. Now every prime ideal of $L \otimes_K M$ contains $p$ by Prop. 2 (V, p. 118).

We may assume that L and M are algebraically disjoint subextensions of an extension $\Omega$ of K (V, p. 116, Th. 5); then p is a prime ideal by Prop. 1 (V, p. 139). If moreover L or M is separable over K, then $L \otimes_K M$ is a reduced ring by the definition of separable extension (V, p. 119, Def. 1); so $p = 0$ and $L \otimes_K M$ is an integral domain because p was prime.

Let $L$ be the field of fractions of $A$. Since $B$ is a regular $K$-algebra, $L \otimes_K B$ is an integral domain, hence so is $A \otimes_K B$, since it is isomorphic to a subring of $L \otimes_K B$.

The proof of a) (resp. c)) is identical to that of part a) (resp. d)) of Prop. 3, V, p. 119, after replacing everywhere « reduced ring » by « integral domain » and « separable algebra » by « regular algebra ». Let us prove b).

Let $A$ and $B$ be two regular $K$-algebras. Let $L$ be an extension of $K$. Since $A$ is regular, the ring $L \otimes_K A$ is an integral domain. By Prop. 2, the ring $(L \otimes_K A) \otimes_K B$ is an integral domain, because $B$ is regular. Finally, the ring $L \otimes_K (A \otimes_K B)$ is isomorphic to $(L \otimes_K A) \otimes_K B$, and hence is an integral domain. This shows $A \otimes_K B$ to be a regular $K$-algebra.

If $L \otimes_K A$ is an integral domain, it may be embedded in its field of fractions $F$. Write $u(x) = x \otimes 1$ for $x \in L$ and denote by $v$ the $K$-homomorphism of $E$ into $F$ which extends the injective homomorphism $y \mapsto 1 \otimes y$ of $A$ into $F$. By Prop. 6 (V, p. 14) the subfields $u(L)$ and $v(E)$ of $F$ are linearly disjoint over $K$; therefore the homomorphism $u * v$ of $L \otimes_K E$ into $F$ (V, p. 12) is injective. This shows $L \otimes_K E$ to be an integral domain.

The condition is necessary by Prop. 4 and sufficient by Prop. 3, a).

By the preceding corollary it is enough to prove that every polynomial algebra $A = K[X_i]_i$, is a regular $K$-algebra. Let $L$ be an extension of $K$; the ring $L \otimes_K A$ is isomorphic to $L[X_i]_{i \in I}$ (III, p. 449, Remark 2), and hence an integral domain (IV, p. 9, Prop. 8).

The first assertion follows from Prop. 3, a).

Let M be an extension of K and let $\mathscr{U}$ be the set of all finitely generated subextensions of L. For each E $\in \mathscr{U}$, the ring $M \otimes_K E$ may be identified with a subring of $M \otimes_K L$ and we thus have an increasing directed family of subrings of $M \otimes_K L$ whose union is $M \otimes_K L$. Now the second assertion follows immediately.

Let E be an extension of K; since L is regular over K, $E \otimes_K L$ is an integral domain. By Prop. 2 (V, p. 141) the ring $(E \otimes_K L) \otimes_L M$ is thus an integral domain and the same is true of the ring $E \otimes_K M$ isomorphic to it (II. p. 278, Prop. 2). Hence the result.

a) If M is regular over K, then the field of fractions of the integral domain $L \otimes_K M$ is a regular extension of L.

b) If L and M are regular extensions of K, the same is true of the field of fractions of $L \otimes_K M$.

Assertion a) follows from Prop. 3, c) (V, p. 141) and the Cor. of V, p. 141; Assertion b) follows from Prop. 3, b) (V, p. 141) and the Cor. of V, p. 141.

a) L is separable over K and K is relatively algebraically closed in L.

b) L is a regular extension of K.

c) The ring $\overline{K} \otimes_K L$ is an integral domain.

d) Let $\overline{L}$ be an algebraic closure of L. Then L is linearly disjoint over K from the relative algebraic closure of K in $\overline{L}$.

Moreover, when these conditions hold, then $\overline{K} \otimes_K L$ is a field.

a) $\Rightarrow$ b): Let M be an extension of K. Under the hypotheses of a), the ring $M \otimes_K L$ is an integral domain, by V, p. 140, Cor.

b) $\Rightarrow$ c): This follows from Def. 2.

c) $\Rightarrow$ d): With the notation of d) we can identify $\overline{K}$ with the relative algebraic closure of K in $\overline{L}$ (V, p. 22, Ex. 2). Suppose that the ring $A = \overline{K} \otimes_K L$ is an integral domain. Let E be a subextension of K of finite degree over K; the subring

E $\otimes_K$ L of A is an integral domain and hence is an algebra of finite degree over L; by the Cor. of V, p. 10 it is a field. Since $\bar{K}$ is the union the increasing directed set of extensions E of the above type, A is a field (V, p. 11, Prop. 3). The canonical homomorphism of A into $\bar{L}$ mapping $x \otimes y$ to $xy$ (for $x \in \bar{K}$ and $y \in L$) is therefore injective, so L and $\bar{K}$ are linearly disjoint over K.

$d) \Rightarrow a)$: Under the hypotheses of d) we have $L \cap \bar{K} = K$, so K is relatively algebraically closed in L; further if p is the characteristic exponent of K, the field $L$ is linearly disjoint from $K^{p^{-\infty}}$ over $K$, hence L is separable over K (V, p. 123, Cor. 1).

The stated condition is clearly necessary. Conversely, assume that $\bar{K} \otimes_K A$ is an integral domain and denote by E the field of fractions of A. By Prop. 4 (V, p. 141) the ring $\bar{K} \otimes_K E$ is an integral domain, hence E is a regular extension of K, by Prop. 9; by V, p. 141, Cor., we conclude that A is regular as K-algebra.

This follows from Cor. 1.

By Cor. 2, A and B are regular K-algebras, and it suffices to apply Prop. 2 (V, p. 141).

Put $w = u * v$ (V, p. 12), denote by F the field of fractions of the integral domain $L \otimes_K M$ and identify L (resp. M) with a subfield of F by means of the mapping $x \mapsto x \otimes 1$ (resp. $y \mapsto 1 \otimes y$); then the restriction of w to L (resp. M) is u (resp. v). Let B be a transcendence basis of M over K (V, p. 109, Th. 1).

By hypothesis $u(L)$ and $v(M)$ are algebraically disjoint over K; therefore (V, p. 114, Prop. 14), $u(L)$ and $v(K(B))$ are linearly disjoint over K. Thus there exists a K-homomorphism $u': L(B) \to E$ which agrees with u on L and with v ou K(B). By construction L and M are linearly disjoint over K in F; by Prop. 8 (V, p. 15) the subfields $L(B)$ and $M$ of F are linearly disjoint over $K(B)$. It follows that there exists a K-homomorphism $w' : M[L(B)] \to E$ which agrees with $u'$ on $L(B)$ and with $v$ on $M$. But the field $F$ is generated by $M \cup L(B)$ and $M$ is algebraic over $K(B)$; we thus have $M[L(B)] = F$ (V, p. 18, Cor. 2). Hence we conclude that $w'$ is a K-isomorphism of $F$ onto $E$ whose restriction to $L$ (resp. $M$) is $u$ (resp. $v$). This shows $u(L)$ and $v(M)$ to be linearly disjoint over $K$.

The ring $L \otimes_K M$ is an integral domain by definition of regular extension and it suffices now to apply Prop. 10.

By Prop. 10 it is enough to remark that the ring $L \otimes_K M$ is an integral domain (V, p. 140, Cor.).

Exercises