The uniqueness of $v$ is clear. Let us put $A = K[x_i]_{i \in I}$ and $A' = K'[x'_i]_{i \in I}$. By hypothesis the monomials $x'' = \prod_{i \in I} x_i^{\alpha_i}$ (for $\alpha = (\alpha_i)_{i \in I}$, in $\mathbf{N}^{(I)}$) form a basis of the vector K-space $A$ and there is a similar property of $A'$. Hence there exists a ring isomorphism $w : A \to A'$ mapping each element $\sum_{\alpha \in \mathbf{N}^{(I)}} c_\alpha x^\alpha$ to $\sum_{\alpha \in \mathbf{N}^{(I)}} u(c_\alpha) {x'}^\alpha$. Since $L$ is the field of fractions of $A$ and $L'$ that of $A'$, the isomorphism $w$ extends to an isomorphism $v$ of the field $L$ onto the field $L'$.

This proposition follows immediately from Def. 1.

b) $S$ is algebraically free over $K$ and $T$ is algebraically free over $K(S)$. c) $T$ is algebraically free over $K$ and $S$ is algebraically free over $K(T)$.

Evidently it is enough to prove that a) and b) are equivalent.

$a) \Rightarrow b)$: Suppose that a) holds. Since $S$ is contained in $S \cup T$, it is algebraically free over $K$. If $T$ is not algebraically free over $K(S)$, there exists (Prop. 3) a finite family $(y_j)_{1 \leq j \leq n}$ of distinct elements of $T$ which is algebraically related over $K(S)$. Hence there is a non-zero polynomial $f$ in the ring $K(S)[Y_1, ..., Y_n]$ such that $f(y_1, ..., y_n) = 0$; after multiplying $f$ if necessary by a non-zero element of $K[S]$ we may suppose that all the coefficients off belong to $K[S]$. The coefficients of fare polynomials in a finite number of distinct elements $x_i \ (1 \leq i \leq m )$ of $S$, with coefficients in $K$. The elements $x_1, ..., x_m, y_1, ..., y_n$ are pairwise distinct because $S \cap T = (21$. The relation $f(y_1, ..., y_n) = 0$ may thus be written

where $g$ is a non-zero polynomial of $K[X_1, ..., X_m, Y_1, ..., Y_n]$, and such a relation contradicts the hypothesis that $S \cup T$ is algebraically free.

$b) \Rightarrow a)$: Suppose that b) holds. In the first place it is clear that $T \cap K(S) = \varnothing$ and $a fortiori \ S \cap T = (21$. It suffices to show that if $x_i \ (1 \leq i \leq m )$ are distinct elements of $S$, finite in number, and $y_j \ (1 \leq j \leq n )$ distinct elements of $T$ finite in number, then the set of the $x_i$ and $y_j$ is algebraically free over $K$ (Prop. 3). Consider a polynomial $f \in K[X_1, ..., X_m, Y_1, ..., Y_n]$ such that $f(x_1, ..., x_m, y_1, ..., y_n) = 0$ and put $f = \sum \varphi_\alpha Y_1^{\alpha_1} ... Y_n^{\alpha_n}$ with $\varphi_\alpha \in K[X_1, ..., X_m]$ for all $\alpha = (\alpha_1, ..., \alpha_n) \in \mathbf{N}^n$. Let $g = f(x_1, ..., x_m, Y_1, ..., Y_n)$; then $g$ is a polynomial in the ring $K[S][Y_1, ..., Y_n]$ and the relation $f(x_1, ..., x_m, y_1, ..., y_n) = 0$ may be written $g(y_1, ..., y_n) = 0$. Since $T$ is algebraically free over $K(S)$, each of the coefficients $\varphi_\alpha(x_1, ..., x_m)$ of $g$ is zero; since $S$ is algebraically free over $K$, we have $\varphi_\alpha = 0$ for all $\alpha \in \mathbf{N}^n$, and hence $f = 0$.

The condition is necessary by Prop. 4.

To prove the sufficiency it is enough (Prop. 3) to show that every finite sequence $(x_1, ..., x_n)$ of distinct elements of $S$ is algebraically free. Now by hypothesis $x_i$ is transcendental over $K(x_1, ..., x_{i-1})$ for $1 \leq i \leq n$, and now our assertion follows by induction on $n$ from the Cor. to Prop. 4.

Arguing by contradiction, let us suppose that X is algebraically related over K'. By Prop. 5 there exists an element $x \in X$ which is algebraic over the field $K'(M)$, where $M = X - \{x\}$. Since $K'(M) = K(M)(K')$ and $K'$ is algebraic over $K$, Cor. 2 of V, p. 18 shows that $K'(M)$ is algebraic over $K(M)$; since $x$ is algebraic over $K'(M)$, it is therefore algebraic over $K(M) = K(X - \{x\})$, by Prop. 3 of V, p. 19. Now Prop. 5 shows X to be algebraically related over K and we have a contradiction.

Let B be a transcendence basis of E over K and $x \in E - B$. Then $x$ is algebraic over $K(B)$; by V, p. 107, Prop. 4, the subset $B \cup \{x\}$ of E is not algebraically free over K, whence the first part of the proposition. Secondly, if E is algebraic over $K(S)$ and B is a maximal algebraically free part of S, it follows from the Cor. of Prop. 4 that every $x \in S$ is algebraic over $K(B)$; hence (V, p. 18, Cor. 1) $K(S)$ is algebraic over $K(B)$, and so (V, p. 19, Prop. 3), E is algebraic over $K(B)$.

By contrast, an extension is not always a pure extension of an algebraic extension (V, p. 171, Ex. 2).

This theorem is a consequence of the following more precise result:

For the set of algebraically free subsets of S, ordered by inclusion, is a set of finite character (E, III, p. 34) by Prop. 3. By Th. 1 of E, III, p. 35 it has a maximal element B containing T, and B is a transcendence basis of E over K, by Prop. 7.

For E is algebraic over K(T U S) and we have T ⊂ T U S.

It is enough to prove the inequality Card(B) ≥ Card(B'), when B and B' are two transcendence bases of E over K, and we may suppose that B' is not empty. We first suppose that B is finite and use induction on its cardinal n; for n = 0, E is algebraic over K and B' is empty. Suppose now that n ≥ 1; given x ∈ B', the exchange theorem provides a subset C of B such that x ∉ C and {x} U C is a transcendence basis of E over K; we have C ≠ B by Prop. 7 whence Card C < n. Put K_1 = K(x) and C' = B' - {x}; then C and C' are algebraically free over the field K_1 (V, p. 107, Prop. 4) and E is algebraic both over K_1(C) = K(C U {x})) and K_1(C') = K(B'). In other words, C and C' are two transcendence bases of E over K_1. Since Card(C) < n, the induction hypothesis implies the inequality Card(C') ≤ Card(C) ≤ n - 1, whence Card(B') ≤ n = Card(B).

Suppose now that B is infinite. Every x ∈ B is algebraic over K(B') and so there exists a finite subset S(x) of B' such that x is algebraic over K(S(x)). Write S = U_{x ∈ B} S(x), then S ⊂ B' and since B is infinite, we have Card(S) ≤ Card(B) (E, III, p. 49, Cor. 3). But since every element of B is algebraic over K(S) and E is algebraic over K(B), we conclude that E is algebraic over K(S) (V, p. 19, Prop. 3). Now Prop. 7 implies that S = B', whence the desired inequality Card(B') ≤ Card(B).

For $F$ is algebraic over $E(T)$; moreover, $E(T)$ is algebraic over the field $K(S \cup T) = K(S)(T)$, because $E$ is algebraic over $K(S)$ ($V$, p. 18, Cor. 2); therefore ($V$, p. 19, Prop. 3) $F$ is algebraic over $K(S \cup T)$. On the other hand, $T$ being algebraically free over $E$ is so $a fortiori$ over $K(S)$, hence ($V$, p. 107, Prop. 4) $S \cup T$ is algebraically free over $K$ and $S \cap T = \varnothing$.

The condition is clearly necessary.

Suppose then that $\Omega$ has the same transcendence degree over $E$ and $F$ and let us choose a transcendence basis $B$ of $\Omega$ over $E$ and a transcendence basis $C$ of $\Omega$ over $F$. Since $B$ and $C$ are equipotent, Prop. 2 ($V$, p. 106) shows that $u$ extends to a $K$-isomorphism $u'$ of $E(B)$ onto $F(C)$. Since $C?$ is an algebraic closure of $E(B)$ and $F(C)$, the Cor. of $V$, p. 23 shows that $u'$ extends to an automorphism $v$ of $\Omega$.

Let us denote by $n, d(E)$ and $d(F)$ the transcendence degrees of $E$ over $K$, of $O$ over $E$ and of $C?$ over $F$ respectively. The existence of the $K$-isomorphism $u$ shows that the transcendence degree of $F$ over $K$ is equal to $n$. By the Cor. of Th. 4 the transcendence degree of $\Omega$ over $K$ is equal to $d(E) + n$ and also to $d(F) + n$. Therefore ($E$, III, p. 28, Prop. 8) we have $d(E) = d(F)$ and we can apply Prop. 8.

Since $\Omega$ is not algebraic over K, there exists an element x of $\Omega$ transcendental over K; then the elements $x^n$ ($n \in \mathbf{N}$) are distinct and transcendental over K. Suppose that x and y are distinct and transcendental over K; by Prop. 2 (V, p. 106) there exists a K-isomorphism $\tilde{u}$ of $K(x)$ onto $K(y)$ such that $\tilde{u}(x) = y$; since $K(x)$ is of transcendence degree 1 over K, Cor. 2 of Prop. 8 shows that $\tilde{u}$ extends to a K-automorphism u of $\Omega$.

a) For x to be algebraic over K it is necessary and sufficient for the set of elements $u(x)$ as u runs over G to be finite,

b) For x to be p-radical over K it is necessary and sufficient that $u(x) = x$ for all $u \in G$.

In particular if K is perfect, the set of invariants of the group G is equal to K.

Suppose first that x is transcendental. By Prop. 9 the set T of elements of $\Omega$ transcendental over K is infinite, and for each $y \in T$ there exists $u \in G$ such that $u(x) = y$. Hence the set of elements $u(x)$ as u runs over G is infinite.

Suppose now that x is algebraic over K and denote by fits minimal polynomial over K; the set of roots of f in $\Omega$ is finite and for each $u \in G$ we have $f(u(x)) = u(f(x)) = 0$. Hence the set of elements $u(x)$ as u runs over G is finite. This proves a).

Let L be the set of elements y of $\Omega$ such that $u(y) = y$ for all $u \in G$, and let $\bar{K}$ be the relative algebraic closure of K in $\Omega$. By what has been said, L is a subextension of $\bar{K}$ over K. Further (Cor. 1 to Prop. 8) every K-automorphism of $\bar{K}$ is the restriction to $\bar{K}$ of an element of G. Assertion b) of Prop. 10 now follows from Cor. 3 of V, p. 53.

This follows from Def. 5.

a) E and F are algebraically disjoint;

b) there exists a transcendence basis of E over K which is algebraically free over F;

c) every subset of E which is algebraically free over K is algebraically free over F.

Let us introduce the following conditions:

$b'$') there exists a transcendence basis of F over K which is algebraically free over E;

$c'$') every subset of F which is algebraically free over K is algebraically free over E.

$a) \Rightarrow b'$') : Suppose that E and F are algebraically disjoint. Let B (resp. C) be a transcendence basis of E (resp. $F$) over K. Then $B \cap C = \varnothing$ and $B \cup C$ is algebraically free over K, hence C is algebraically free over $K(B)$ (V, p. 107, Prop. 4); since E is algebraic over $K(B)$, Prop. 6 of V, p. 108 shows C to be algebraically free over E.

$b'$') $\Rightarrow$ c) : Suppose that there exists a transcendence basis C of F over K which is algebraically free over E. Let A be a subset of E algebraically free over K. Then C is algebraically free over $K(A)$, hence A is algebraically free over $K(C)$ (V, p. 107, Prop. 4) and therefore over $F$ (V, p. 108, Prop. 6), because F is algebraic over $K(C)$.

c) $\Rightarrow$ a) : This follows at once from Prop. 4 (V, p. 107).

The implications $a) \Rightarrow b) \Rightarrow c' ) \Rightarrow a)$ may be proved in the same way.

Let B be a transcendence basis of E over K; this is also a transcendence basis of E' over K. Since E is algebraically disjoint from F over K, B is algebraically free over F, hence over F' (V, p. 108, Prop. 6); now we can apply Prop. 12.

a) We have $\operatorname{tr. deg}_L F(E) \leq \operatorname{tr. deg}_K E$. When E and F are algebraically disjoint over K, then every transcendence basis of E over K is a transcendence basis of

F(E) over F and we have tr . deg_F F(E) = tr . deg_K E. Conversely, this equality implies that E and F are algebraically disjoint over K when tr . deg_K E is finite.

b) We have tr . deg_K K(E U F) ≤ tr . deg_K E + tr . deg_K F. When E and F are algebraically disjoint over K, we have tr . deg_K K(E U F) = tr . deg_K E + tr . deg_K F. Conversely, this equality implies that E and F are algebraically disjoint over K when E and F are of finite transcendence degree over K.

a) Let B be a transcendence basis of E over K; then E is algebraic over K(B), and Cor. 2 of V, p. 18 shows that F(E) is algebraic over F(K(B)) = F(B). By Th. 2 (V, p. 109) B contains a transcendence basis of F(E) over F; when E is algebraically disjoint from F over K, B is algebraically free over F (Prop. 12) and this is then a transcendence basis of F(E) over F. The three first assertions of a) follow from this. Suppose now that E is of finite transcendence degree over K, equal to that of F(E) over F; since F(E) is algebraic over F(B) and Card B = tr . deg_F F(E), Cor. 1 of V, p. 110 shows B to be algebraically free over F, so E is algebraically disjoint from F over K (Prop. 12).

b) We have K(E U F) = F(E) and so the Cor. of V, p. 110 implies the equality:

Now b) follows at once from a) and this equality.

For E and F to be algebraically disjoint over K it is necessary and sufficient for B to be algebraically free over F (Prop. 12), that is, for the monomials in the elements of B to be linearly independent over F. Since these monomials form a basis of the vector K-space K[B], it comes to the same to say that K[B] and F are linearly disjoint over K. Finally, since K(B) is the field of fractions of K[B], Prop. 6 of V, p. 14 shows that K[B] and F are linearly disjoint if and only if this is so for K(B) and F.

a) The family $(E_i)_{i \in I}$ is algebraically free.

b) For each $i \in I$ the extension $E_i$ is algebraically disjoint over $K$ from the extension $F_i$ generated by the $E_j$ for $j \neq i$.

c) There exists a family $(B_i)_{i \in I}$ of disjoint subsets of $L$, such that $B_i$ is a transcendence basis of $E_i$ over $K$ for each $i \in I$, and $B = \bigcup_{i \in I} B_i$ is algebraically free over $K$.

It is clear that a) implies c).

Assuming c), let us choose $i$ in $I$; put $C_i = \bigcup_{j \neq i} B_j$. For each $j \neq i$ every element of $E_j$ is algebraic over $K(B_j)$ and a fortiori over $K(C_i)$. By Cor. 1 of $V$, p. 18, the field $F_i$ is therefore algebraic over $K(C_i)$. Further, we have $B_i \cap C_i = \varnothing$ and $B - B_i \cup C_i$ is algebraically free over $K$; therefore $B_i$ is algebraically free over $K(C_i)$ ($V$, p. 107, Prop. 4), hence also over $F_i$ (which is algebraic over $K(C_i)$) by Prop. 6 of $V$, p. 108. Thus we have proved that $E_i$ is algebraically disjoint from $F_i$ over $K$ ($V$, p. 113, Prop. 12), hence c) implies b).

Let us now assume b) and prove a). It is enough to show that if $i_1, \ldots, i_n$ are distinct elements of $I$, then the family of extensions $(E_{i_1}, \ldots, E_{i_n})$ is algebraically free; we argue by induction on $n$, the case $n = 1$ being trivial. Suppose then that $n > 1$ and that the family $(E_{i_1}, \ldots, E_{i_{n-1}})$ is algebraically free; for $1 \leq k \leq n$ choose a subset $A_k$ of $E_{i_k}$ algebraically free over $K$ and put

B = A, U ... U A_{n-1}. By the induction hypothesis the subsets A_1, ..., A_{n-1} are pairwise disjoint and B is algebraically free over K; by b) E_{i_n} is algebraically disjoint from F_i and since B is contained in F_{i_n} we have B \cap A_i = \varnothing and B \cup A_i = A, U ... U A_i is algebraically free over K. We have thus shown that the family (E_{i_1}, ..., E_{i_n}) is algebraically free.

The following proposition generalizes part b) of Prop. 13 (V, p. 113).

a) We have tr . deg_K E \leq \sum_{i \in I} \text{tr . deg}_K E_i, with equality when the family (E_i)_{i \in I} is algebraically free over K.

b) Conversely, assume that tr . deg_K E = \sum_{i \in I} \text{tr . deg}_K E_i and that tr . deg_K E is finite; then the family (E_i)_{i \in I}, is algebraically free over K.

For each $i \in I$ let B_i be a transcendence basis of E_i over K and put $B = \bigcup_{i \in I} B_i$. For each $i \in I$, each element of E_i is algebraic over K(B_i), hence over K(B). Now Cor. 1 of V, p. 18 shows that E is algebraic over K(B); by V, p. 109, Th. 2, B thus contains a transcendence basis of E over K. If moreover the family (E_i)_{i \in I} is algebraically free over K, then the B_i are disjoint and the set B is algebraically free over K. This establishes a) (Set Theory, III, p. 160, Cor. of Prop. 4).

Under the hypotheses of b), E is algebraic over K(B) and of finite transcendence degree over K, and we have Card(B) \leq \text{tr . deg}_K E. By Cor. 1 of V, p. 110, B is algebraically free over K and the B_i are disjoint. Now Prop. 15 shows that the family (E_i)_{i \in I} is algebraically free over K.

Before stating the next theorem let us remark that there exist algebraically closed extensions of K of arbitrary transcendence degree, for example an algebraic closure of an appropriate field of rational fractions.

holds. Then there exists an algebraically free family (F_i)_{i \in I}, of subextensions of \Omega such that F_i is K-isomorphic to E_i for all i \in I.

For each $i \in I$ let B_i be a transcendence basis of E_i over K. Let B be a transcendence basis of \Omega over K. By (2) we have Card B \geq \sum_{i \in I} \text{Card} B_i; hence there exists a family (B'_i)_{i \in I}, of pairwise disjoint subsets of B and bijections u_i : B_i \to B'_i for i \in I. By Prop. 2 of V, p. 106, u, extends to a K-isomorphism v_i of K(B_i) onto K(B'_i); since \Omega is algebraically closed and E_i algebraic over

K(B_i), the Cor. (V, p. 23) shows that $v_i$ extends to a K-isomorphism of $E_i$ onto a subextension $F_i$ of $\Omega$. By construction $B'_i$ is a transcendence basis of $F_i$ over K, and Prop. 15 (V, p. 115) shows that the family $(F_i)_{i \in I}$ of subextensions of $\Omega$ is algebraically free over K.

For each field is the ascending directed union of finitely generated subfields over its prime field, and now it is enough to apply Cor. 1. \* Example. — This applies particularly in characteristic 0 in taking $\Omega = \mathbf{C}$ (« Lefschetz' principle »). \*

Suppose that E is finitely generated over K and let S be a finite subset of E such that $E = K(S)$. By Th. 2 (V, p. 109) S contains a transcendence basis $B'$ of E over K and this has the same cardinal as B (V, p. 110, Th. 3). Therefore B is finite. Put $K' = K(B)$; then E is algebraic over $K'$ and we have $E = K'(S)$; since S is finite, Th. 2 of V, p. 18 shows that $[E : K']$ is finite.

Conversely, suppose that B is finite and $[E : K(B)]$ is finite. If C is a (finite) basis of the vector space E over $K(B)$, we have $E = K(B)(C) = K(B \cup C)$ and so E is a finitely generated extension of K.

Let B be a transcendence basis of E over K. By Cor. 2 of V, p. 114, $K'$ is linearly disjoint from $K(B)$ over K, whence $[K' : K] = [K'(B) : K(B)] \leq [E : K(B)]$ and now the finiteness of $[K' : K]$ follows from that of $[E : K(B)]$.

By Cor. 1 we are reduced to proving that a field L which is an extension of finite degree of its prime subfield possesses only a finite number of roots of unity. This is clear when L is of characteristic $\neq 0$ because it is then finite. If L is of characteristic 0 and contains infinitely many roots of unity, then it contains primitive roots of unity of arbitrarily high order. By V, p. 84, Th. 2 there exist therefore an infinity of integers $n > 0$ such that $\varphi(n) \leq [L : \mathbf{Q}]$, which is absurd (V, p. 80, Formulae (2) and (3)).

Let B' be a transcendence basis of E' over K. By V, p. 109, Th. 2, B' is contained in a transcendence basis B of E over K and so is finite, by Prop. 17. Since E' is algebraic over K(B') and E is a finitely generated extension of K(B'). Cor. 1 shows $[E' : K(B')]$ to be finite. Now Prop. 17 shows that E' is finitely generated over K.

Prop. 17 may be paraphrased by saying that a finitely generated extension of K is an algebraic extension of finite degree of a purely transcendental extension $K(x_1, \ldots, x_n)$.