Note first that it is sufficient to prove the theorem when A is a polynomial algebra $k[Y_1, \ldots, Y_m]$. For in the general case, A is isomorphic to a quotient of such an algebra A' by an ideal $a'_0$; let $a'_j$ denote the inverse image of $a_j$ in A' and let $x'_i$ ($1 \leq i \leq r$) be elements of A' satisfying the conditions of the statement for the ring A' and the increasing sequence of ideals $a'_0 \subset a'_1 \subset \ldots \subset a'_p$. Then the images $x_i$ of the $x'_i$ in A for $i > h(0)$ satisfy the desired conditions; this is obvious for condition (b) and for condition (a) this follows from § 1, no. 1, Proposition 2; finally, if the $x_i(h(0) + 1 \leq i \leq r)$ were not algebraically independent over k, there would be a non-zero polynomial

where the $Q_j$ are polynomials in $k[Y_1, \ldots, Y_m]$; as $a_p \neq 0$ is invertible in $k$, (4) is certainly an equation of integral dependence with coefficients in B, whence our assertion.

The field of fractions $k(Y_1, \ldots, Y_r)$ of A is therefore algebraic over the field of fractions $k(x_1, \ldots, x_m)$ of B, which proves (Algebra, Chapter V, § 5, no. 3, Theorem 4) that the $x_i$ ($1 \leq i \leq m$) are algebraically independent. Moreover, $a \cap B = Bx_1$; for every element $z \in a, \cap B$ may be written $z = x_1 z'$ where $z' \in A \cap k(x_1, \ldots, x_r)$; but $A \cap k(x_1, \ldots, x_r) = k[x_1, \ldots, x_m] = B$ since $B$ is integrally closed ($\S$ 1, no. 3, Corollary 2 to Proposition 13); therefore $z' \in B$, which completes the proof of properties (a) and (b) in this case.

(A2) General case ($p = 1$). We argue by induction on $m$, the case $m = 0$ being trivial. We may obviously suppose that $a_i \neq 0$ (otherwise we may take $x_i = Y$, for $1 \leq i \leq m$ and $h(1) = 0$). Let $x_1$ be a non-zero element of $a_i$; by (A1) there exist $t_2, \ldots, t_m$ such that $x_1, t_2, \ldots, t$, are algebraically independent over $k$, $A$ is integral over $C = k[x_1, t_2, \ldots, t_m]$ and $x_1 A \cap C = x_1 C$. By the induction hypothesis there exist elements $x_2, \ldots, x_m$ of $k[t_2, \ldots, t_m]$ and an integer $h$ such that $k[t_2, \ldots, t_m]$ is integral over $B' = k[x_2, \ldots, x_m]$, $x_2, \ldots, x_m$ are algebraically independent over $k$ and the ideal $a_i \cap B'$ is generated by $x_2, \ldots, x_h$. Then $C$ is integral over $B = k[x_1, x_2, \ldots, x_m]$ ($\S$ 1, no. 1, Corollary to Proposition 5 ) and hence so is $A$ ($\S$ 1, no. 1, Proposition 6 ); the same argument as in the case (A1) shows that $x_1, \ldots, x_m$ are algebraically independent over $k$; finally, as $x_1 \in a_i$, and $B = B'[x_1], a_i \cap B = Bx, + (a_i \cap B')$ and, as $a_i \cap B'$ is generated (in $B'$) by $x_2, \ldots, x_h, a_i \cap B$ is generated (in $B$) by $x_1, x_2, \ldots, x_h$.

(B) Passage from $p - 1$ top. Let $t_1, \ldots, t_r$, be elements of $A$ satisfying the conditions of the theorem for the increasing sequence of ideals $a_1 \subset \ldots \subset a_{p-1}$ and let us write $r = h(p - 1)$. By (A2) there exist elements $x_{r+1}, \ldots, x_s$ of $k[t_{r+1}, \ldots, t_m]$ and an integer s such that

is integral over $B' = k[x_{r+1}, \ldots, x_m]$, $x_{r+1}, \ldots, x_s$ are algebraically independent over $k$ and the ideal $a_p \cap B'$ is generated by $x_{r+1}, \ldots, x_s$. Writing $x_i = t_i$ for $i \leq r$ the family $(x_i)_{1 \leq i \leq m}$ obtained solves the problem with $h(p) = s$. For $A$ is integral over $C[t_1, \ldots, t_r] = C[x_1, \ldots, x_r]$ and hence also over $B = k[x_1, \ldots, x_m] = B'[x_1, \ldots, x_r]$ since $C$ is integral over $B'$ ($\S$ 1, no. 1, Corollary to Proposition 5 and Proposition 6); it can be shown as in the case (A1) that the $x_i$ are algebraically independent over $k$. On the other hand, for $j \leq p - 1$, the ideal is by hypothesis the set of polynomials in $x_1, \ldots, x_r, t_{r+1}, \ldots, t_m$ all of whose monomials contain one of the elements $x_1, \ldots, x_{h(j)}$; as $x_{r+1}, \ldots, x_m$ are polynomials in $t_{r+1}, \ldots, t$, with coefficients in $k$, it is seen immediately that a polynomial in $x_1, \ldots, x_r, x_{r+1}, \ldots, x_s$ (with coefficients in $k$) can belong to $a_j$ only if all its monomials contain one of the elements $x_1, \ldots, x_{h(j)}$. Finally, as $x_1, \ldots, x_r$ belong to $a_{p-1}$, and hence also to $a_p$, $a_p \cap B'[x_1, \ldots, x_r]$ consists of the polynomials in $x_1, \ldots, x_r$ with coefficients in $B'$ whose constant term belongs to $a_p \cap B'$; this ideal is therefore generated by $x_1, \ldots, x_r, x_{r+1}, \ldots, x_t$.

We write $S = A - \{0\}$ and let $k = S^{-1}A$ the field of fractions of $A$; clearly $S^{-1}B$ is a finitely generated $k$-algebra and, as it contains $k$ by hypothesis (Chapter II, § 2, no. 4, Theorem 1), it is not reduced to 0. By Theorem 1 (applied top = 1 and $a_i = 0$) there exists therefore a finite sequence $(x_i)_{1 \leq i \leq n}$ of elements of $S^{-1}B$ which are algebraically independent over $k$ and such that $S^{-1}B$ is integral over $k[x_1, \ldots, x_n]$. Let $(z_j)_{1 \leq j \leq m}$ be a system of generators of the $A$-algebra $B$; in $S^{-1}B$ each of the $z_j/1$ satisfies an equation of integral dependence

where the $P_{hj}$ are polynomials in the $x_i$ with coefficients in $k$. There exists an element $s \neq 0$ of $A$ such that we may write $x_i = y_i/s$ whereby, $\in B$ for $1 \leq i \leq n$ and all the coefficients of the $P_{hj}$ are of the form $c/s$ where $c \in A$; finally, replacing if need be, $s$ by a product of elements of $S$, we may assume that in $B$

where the $Q_{hj}$ are polynomials in $y_1, \ldots, y_n$ with coefficients in $A$; if we write $z'_j = sz_j$, it is seen, multiplying (6) by $s^{a_j-1}$, that $z'_j$ is integral over $B' = A[y_1, \ldots, y_n]$. We show that the $y_i$ are algebraically independent over $A$; if there is a relation of the form $\sum_p a_p y_1^{p_1} \ldots y_n^{p_n} = 0$ where $a_p \in A$ for all $p$, we deduce that $\sum_p a'_p x_1^{p_1} \ldots x_n^{p_n} = 0$ in $S^{-1}B$, where $a'_p = a_p s^{p_1 + \cdots + p_n}$ in $k$; by hypothesis therefore $a'_p = 0$ for all $p$, whence $a_p = 0$ for all $p$. Moreover, in the ring $B[s^{-1}]$ each of the $z'_j/1$ is integral over $B'[s^{-1}]$ (§ 1, no. 1, Proposition 2) and, as $z_j/1 = (z'_j/1)(1/s)$ in $B[s^{-1}]$, it is seen that the $z_j/1$ are integral over $B'[s^{-1}]$, which completes the proof (§ 1, no. 1, Proposition 4).

It follows from Corollary 1 that there exist elements $x_1, \ldots, x_n$ of $K$ and an element $a \neq 0$ of $A$ such that $x_1, \ldots, x_n$ are algebraically independent over $A$ (and therefore over L) and that K is integral over the subring $A[x_1, \ldots, x_n, a^{-1}]$. Then it follows from §2, no.1, Lemma 2 that $A[x_1, \ldots, x_n, a^{-1}]$ is a field. But the only invertible elements of a polynomial ring $C[Y_1, \ldots, Y_n]$ over an integral domain C are the invertible elements of C; applying this remark to $C = A[a^{-1}]$, it is seen that necessarily $n = 0$ and that $A[a^{-1}]$ is a field equal to L by definition of the latter. As K is integral over L and is a finitely generated L-algebra, the degree [K:L] is finite (§ 1, no. 1, Proposition 4).

The hypothesis on b implies that, if h is the canonical homomorphism $x \mapsto x/1$ of B to $B[b^{-1}]$, the homomorphism $h \circ \varphi$ of A to $B[b^{-1}]$ is injective. By Corollary 1 there therefore exist an element $a \neq 0$ of A and a subring $B'$ of $B[b^{-1}]$ such that $B[b^{-1}, a^{-1}]$ is integral over $B'[a^{-1}]$ and $B'$ is isomorphic to a polynomial algebra $A[Y_1, \ldots, Y_n]$. Let f be a homomorphism from A to an algebraically closed field L such that $f(a) \neq 0$; there exists a homomorphism from $A[Y_1, \ldots, Y_n]$ to L extending f and hence there exists a homomorphism $f'$ from $B'$ to L extending f. As $f'(a) \neq 0$ in L, there exists a homomorphism $f''$ from $B'[a^{-1}]$ to L such that

for all $x \in B'$ and all $n > 0$ (Chapter II, §2, no. 1, Proposition 1). Finally, as $B[b^{-1}, a^{-1}]$ is integral over $B'[a^{-1}]$, there exists a homomorphism $f'''$ from $B[b^{-1}, a^{-1}]$ to L extending $f''$ (§2, no. 1, Corollary 4 to Theorem 1). If $j : x \mapsto x/1$ is the canonical homomorphism from B to $B[b^{-1}, a^{-1}]$, $g = f''' \circ j$ solves the problem for $j(b)$ is invertible in $B[b^{-1}, a^{-1}]$ and hence $f'''(j(b)) \neq 0$ in L.

Note that, if B is assumed to be an integral domain and $A \subset B$ in Corollary 3, the hypothesis on b is equivalent to $b \neq 0$.

By Theorem 1 there exists a subalgebra C of A isomorphic to a polynomial algebra $k[X_1, \ldots, X_n]$ and such that A is integral over C; $A'$ is obviously the integral closure of C in K' (§ 1, no. 1, Proposition 6); we may therefore confine our attention to the case where $A = k[X_1, \ldots, X_n]$. Let $N$ be the quasi-Galois extension of $K$ (in an algebraic closure of $K$) generated by $K'$, which is a finite algebraic extension of $K$ (Algebra, Chapter V, § 6, no. 3, Corollary 1 to Proposition 9). It will suffice to prove that the integral closure $B$ of $A$ in $N$ is a finitely generated $A$-module, for $A'$ is a sub-$A$-module of $B$ and $A$ is a Noetherian ring (Chapter III, § 2, no. 10, Corollary 2 to Theorem 2). We may therefore confine our attention to the case where $K'$ is a quasi-Galois extension of $K$. Then we know (Algebra, Chapter V, § 10, no. 9, Proposition 14) that $K'$ is a (finite) Galois extension of a (finite) radicial extension $K''$ of $K$. If $A''$ is the integral closure of $A$ in $K''$, $A'$ is the integral closure of $A''$ in $K'$ and it will suffice to prove that $A''$ is a finitely generated $A$-module and $A'$ is a finitely generated $A''$-module. Now, if it has been proved that $A''$ is a finitely generated $A$-module, it is a Noetherian domain, integrally closed by definition; the fact that $A$ is a finitely generated $A''$-module will follow from § 1, no. 6, Corollary 1 to Proposition 18.

We see therefore that we may confine our attention to the case where $A = k[X_1, \ldots, X_n]$ and where $K'$ is a finite radicial extension of $K = k(X_1, \ldots, X_r)$. Then $K'$ is generated by a finite family of elements $(y_i)_{1 \leq i \leq m}$ and there exists a power $q$ of the characteristic exponent of $k$ such that $y_i^q \in k(X_1, \ldots, X_r)$. Let $c_j$ ($1 \leq j \leq r$) be the coefficients of the numerators and denominators of the rational functions in $X_1, \ldots, X_r$, equal to $y_i^q$ ($1 \leq i \leq m$). Then $K'$ is contained in the extension $L = k'(X_1^{q^{-1}}, \ldots, X_n^{q^{-1}})$, where $k' = k(c_1^{q^{-1}}, \ldots, c_r^{q^{-1}})$ (we are in an algebraic closure of $K'$) and $A$ is contained in the algebraic closure $B'$ of $A$ in $L'$. Now, $k'$ is algebraic over $k$ and hence $C' = k'[X_1, \ldots, X_n]$ is integral over $A$ ($§ 1$, no. 1, Proposition 5); as $k'[X_1^{q^{-1}}, \ldots, X_n^{q^{-1}}]$ is integrally closed ($§ 1$, no. 3, Corollary 2 to Proposition 13), it is seen that this ring is the integral closure of $C'$ in $L'$ and hence also that of $A$ ($§ 1$, no. 1, Proposition 6), in other words $B' = k'[X_1^{q^{-1}}, \ldots, X_n^{q^{-1}}]$. Now clearly $B'$ is a finitely generated $C'$-module ($§ 1$, no. 1, Proposition 4) and, as $k'$ is a finite extension of $k$, $C'$ is a finitely generated $A$-module and hence $B'$ is a finitely generated $A$-module; since $A$ is Noetherian and $A \subset B'$, $A'$ is a finitely generated $A$-module.

(i) If $A \neq \{0\}$, there exists a $k$-homomorphism from $A$ to $L$.

(ii) Let $f_1, f_2$ be two $k$-homomorphisms from $A$ to $L$. For $f_1$ and $f_2$ to have the same kernel, it is necessary and sufficient that there exist a $k$-automorphism $s$ of $L$ such that $f_2 = s \circ f_1$.

(iii) Let $a$ be an ideal of $A$. For $a$ to be maximal, it is necessary and sufficient that it be the kernel of a k-homomorphism from $A$ to $L$.

(iv) For an element $x$ of $A$ to be such that $f(x) = 0$ for every k-homomorphism $f$ from $A$ to $L$, it is necessary and sufficient that $x$ be nilpotent.

Assertion (i) follows from no. 1, Corollary 3 to Theorem 1 applied replacing $A$ by $k$, $B$ by $A$, $b$ by the unit element of $B$ and $f$ by the canonical injection of $k$ into $L$.

If $f$ is a k-homomorphism from $A$ to $L$, $f(A)$ is a subring of $L$ containing $k$; as $L$ is an algebraic extension of $k$, $f(A)$ is a field (Algebra, Chapter V, § 3, no. 2, Proposition 3) and, if $a$ is the kernel off, $A/a$, isomorphic to $f(A)$, is therefore a field, which proves that $a$ is maximal. Conversely, if $a$ is a maximal ideal of $A$, it follows from (i) that there exists a k-homomorphism from $A/a$ to $L$ and hence a k-homomorphism of $A$ to $L$ whose kernel $b$ contains $a$; but as $a$ is maximal, $b = a$; this proves (iii).

We now prove (ii). If $s$ is a k-automorphism of $L$ such that $f_2 = s \circ f_1$, clearly $f_1$ and $f_2$ have the same kernel. Conversely, suppose that $f_1$ and $f_2$ have the same kernel; then there exists a k-isomorphism $s_0$ of the field $f_1(A)$ onto the field $f_2(A)$ such that $f_2 = s_0 \circ f_1$; but by Algebra, Chapter V, § 6, no. 3, Proposition 7, $s_0$ extends to a k-automorphism of $L$ and hence $f_2 = s \circ f_1$.

Finally, if $x \in A$ is such that $x^n = 0$, for every k-homomorphism $f$ from $A$ to $L$, $(f(x))^n = f(x^n) = 0$ and hence $f(x) = 0$ since $L$ is a field. Conversely, suppose that $x \in A$ is not nilpotent; then $A[x^{-1}]$ is a finitely generated $A$-algebra (and therefore a finitely generated k-algebra) not reduced to 0 (Chapter 11, § 2, no. 1, Remark 3) and hence there exists a k-homomorphism $g$ from $A[x^{-1}]$ to $L$ by (i). If $j : A \to A[x^{-1}]$ is the canonical homomorphism, $f = g \circ j$ is a k-homomorphism from $A$ to $L$ and $f(x)g(1/x) = g(x/1)g(1/x) = g(1) = 1$, whence $f(x) \neq 0$.

Let $k$ be a field and $L$ an extension field of $k$; an element $x = (x_1, \ldots, x_n)$ of $L^n$ is called a zero in $L^n$ of an ideal $c$ of the polynomial ring $k[X_1, \ldots, X_n]$ if

for all $P \in r$.

There exists a unique k-algebra homomorphism $h$ of $k[X_1, \ldots, X_n]$ onto $A$ such that $h(X_i) = a_i$, for $1 \leq i \leq n$ and by definition $r$ is the kernel of $h$. The mapping f $\mapsto$ his a bijection of the set of k-homomorphisms from $A$ to $L$ onto the set of k-homomorphisms from $k[X_1, \ldots, X_n]$ to $L$ which are zero on c. For every polynomial $P \in k[X_1, \ldots, X_n]$ and every element $x = (x_1, \ldots, x_i) \in L^n$ we write $h_x(P) = P(x)$; then the mapping $x \mapsto h_x$ is a bijection of $L^n$ onto the set of k-homomorphisms from $k[X_1, \ldots, X_n]$ to $L$ (such a homomorphism being determined by its values at the $X_i$ ($1 \leq i \leq n$) ); to say that $h_x$ is zero on c means that $x$ is a zero of c in $L''$, whence the lemma.

If Proposition 1 is applied to the algebra $A = k[X_1, \ldots, X_n]/r$, where c is an ideal of $k[X_1, \ldots, X_n]$ distinct from the whole ring, we obtain by Lemma 1 the following statement:

(i) Every ideal $r \not\subset k[X_1, \ldots, X_n]$ not containing 1 admits at least one zero in $L^n$.

(ii) Let $x = (x_1, \ldots, x_n), y = (y_1, \ldots, y_n)$ be two elements of $L^n$; for the set of polynomials of $k[X_1, \ldots, X_n]$ zero at x to be identical with the set of polynomials of $k[X_1, \ldots, X_n]$ zero at y, it is necessary and sufficient that there exists a k-automorphism s of L such that $y_i = s(x_i)$ for $1 \leq i \leq n$.

(iii) For an ideal a of $k[X_1, \ldots, X_n]$ to be maximal, it is necessary and sufficient that there exist an x in $L^n$ such that a is the set of polynomials of $k[X_1, \ldots, X_n]$ zero at x.

(iv) For a polynomial Q of $k[X_1, \ldots, X_n]$ to be zero on the set of zeros in $L''$ of an ideal c of $k[X_1, \ldots, X_n]$, it is necessary and sufficient that there exist an integer m > 0 such that $Q^m \in c$.

The Jacobson radical (resp. nilradical) of $A/a$ is the intersection of the maximal (resp. prime) ideals of $A/a$ (Algebra, Chapter VIII, § 6, no. 3, Definition 3 and Commutative Algebra, Chapter II, § 2, no. 6, Proposition 13). The stated condition means therefore that for every ideal $a$ of $A$ the intersection of the prime ideals containing $a$ is equal to the intersection of the maximal ideals containing $a$. This condition obviously holds for every ideal $a$ of $A$ if $A$ is a Jacobson ring; conversely, if it holds for every prime ideal of $A$, $A$ is a Jacobson ring by definition.

It is sufficient to note that $A/a$ is a Jacobson ring.

The maximal ideals of $A$ are the $A\mathbf{p}_\lambda$ (loc. cit., no. 2, Proposition 2). If $L$ is finite, their intersection is the ideal $Ax$, where $x = \prod_{\lambda \in L} \mathbf{p}_\lambda$ (ibid.) and hence different from $(0)$; on the other hand, if $L$ is infinite, the intersection of the $A\mathbf{p}_\lambda$ is $(0)$, every element $\neq 0$ of $A$ being divisible by only a finite number of extremal elements (loc. cit., no. 3, Theorem 2). The proposition then follows from the fact that $(0)$ is the only prime ideal which is not maximal in $A$ (Algebra, Chapter VI, § 1, no. 13, Proposition 14 (DIV)).

Replacing $A$ by its canonical image in $B$, we may assume that $A \subset B$. Let $\mathfrak{p}'$ be a prime ideal of $B$ and let $\mathfrak{p} = A \cap \mathfrak{p}'$. There exists by hypothesis a family $(\mathfrak{m}_\lambda)_{\lambda \in L}$ of maximal ideals of $A$ whose intersection is equal top. For all $\lambda \in L$ there exists a maximal ideal $\mathfrak{m}'_\lambda$ of $B$ lying above $\mathfrak{m}_\lambda$ and containing $\mathfrak{p}'$ (§ 2, no. 1, Proposition 1 and Corollary 2 to Theorem 1). If we write $\mathfrak{q}' = \bigcap_{\lambda \in L} \mathfrak{m}'_\lambda$, then $\mathfrak{q}' \cap A = \bigcap_{\lambda \in L} \mathfrak{m}_\lambda = \mathfrak{p}$ and $\mathfrak{q}' \supset \mathfrak{p}'$, whence $\mathfrak{q}' = \mathfrak{p}'$ ($2$, no. 1, Corollary 1 to Proposition 1).

Let $\mathfrak{p}'$ be a prime ideal of $B$ and $\mathfrak{p} = \varphi^{-1}(\mathfrak{p}')$. Let $v$ be an element $\neq 0$ of $B/\mathfrak{p}'$.

As $B/p'$ is a finitely generated integral (Alp)-algebra and the canonical homomorphism $\phi : A/p \to B/p'$ is injective, there exists an element $u \neq 0$ of $A/p$ such that, for every homomorphism $f$ from $A/p$ to an algebraically closed field $L$ whose kernel does not contain $u$, there exists a homomorphism $g$ from $B/p'$ to $L$ whose kernel does not contain $v$ and for which $f = g \circ \phi$ (no. 1, Corollary 3 to Theorem 1). Since $A$ is a Jacobson ring, there exists a maximal ideal $m$ of $A$ containing $p$ and such that $u \notin m/p$. We take $L$ to be an algebraic closure of $A/m$ and $f$ to be the canonical homomorphism $A/p \to L$; let

be a homomorphism such that $f = g \circ \phi$ and $g(v) \neq 0$. Then

hence $g(B/p')$ is a subfield of $K$ (Algebra, Chapter V, § 3, no. 2, Proposition 3) and the kernel of $g$ is therefore a maximal ideal of $B/p'$ not containing $v$. Thus it is seen that the intersection of the maximal ideals of $B/p'$ is reduced to 0, which proves that $B$ is a Jacobson ring. Moreover, if $p'$ is maximal, $g$ is necessarily injective and hence $p = m$ is maximal; finally $B/p'$ is then a finitely generated algebra over the field $A/m$ and hence is a finite extension of $A/m$ (no. 1, Corollary 2 to Theorem 1).

If the integral domain $A/p$ is finite, it is a field, as, for every $u \neq 0$ in $A/p$, the mapping $v \mapsto uv$ of $A/p$ to itself is injective and hence bijective since $A/p$ is finite. Conversely, for every maximal ideal $m$ of $A$, the inverse image of $m$ in $A$ is a maximal ideal ($p$) and $A/m$ is finite over the prime field $\mathbf{Z}/(p) = \mathbf{F}_p$ by Theorem 3.