Let $p$ be a prime ideal of $\mathbf{A}$ such that $p \cap S = \varnothing$. Then $a' \mapsto S^{-1}a'$ is a surjective mapping of the set $\mathscr{F}$ of ideals of $\mathbf{A}'$ lying above $p$ onto the set $\mathscr{F}_1$ of ideals of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$ and the mapping $a'_1 \mapsto i'(a'_1)$ is a bijection of $\mathscr{F}_1$ onto the set of ideals belonging to $\mathscr{F}$ and is saturated with respect to $h(S)$; in particular $p' \mapsto S^{-1}p'$ is a bijection of the set of prime ideals of $\mathbf{A}'$ lying above $p$ onto the set of prime ideals of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$.

We know that $S^{-1}p$ is a prime ideal of $S^{-1}\mathbf{A}$ and that $i_1(S^{-1}p) = p$ (Chapter 11, § 2, no. 5, Proposition 11); if there exists an ideal $b'$ of $S^{-1}\mathbf{A}'$ lying above $S^{-1}p$, then $h(i_1(b')) = i(h_1(b')) = p$; as $S^{-1}i'(b') = b'$ (loc. cit.), this already shows that the image of $\mathscr{F}$ under the mapping $a' \mapsto S^{-1}a'$ contains $\mathscr{F}_1$. On the other hand, if $a' \in \mathscr{F}$, $a \in A$ and $s \in S$, there are the following equivalences

$\iff$ there exists $t \in S$ such that $h(t)h(a) \in a'$ $\iff$ there exists $t \in S$ such that $ta \in \mathfrak{p}$ $\iff a/z \in S^{-1}\mathfrak{p}$.

Hence $h_1^{-1}(S^{-1}a') = S^{-1}\mathfrak{p}$, which completes the proof that the image of $\mathscr{F}$ under the mapping $a' \mapsto S^{-1}a'$ is equal to $\mathscr{F}_1$; the other assertions follow from Chapter 11, § 2, no. 5, Proposition 11.

Let us write $B = A/\mathfrak{p}$, $B' = A'/\mathfrak{p}'$ and let $h_1 : B \to B'$ be the homomorphism derived from $h$ by taking quotients; $B$ and $B'$ are integral domains and $B'$ is integral over $B$ ($§ 1$, no. 1, Proposition 2). To say that $\mathfrak{p}$ (resp. $\mathfrak{p}'$) is maximal means that $B$ (resp. $B'$) is a field. The proposition then follows from the following lemma:

If $A$ is a field, then, for ally $\neq 0$ in $B$, $A[y]$ is by hypothesis ($§ 1$, Theorem 1) a finitely generated $A$-module; as $A[y]$ is an integral domain, it is a field (Algebra, Chapter V, § 2, no. 1, Proposition 1) and a fortiori $y$ is invertible in $B$ and hence $B$ is a field. Conversely, suppose that $B$ is a field and let $z \neq 0$ in $A$; as $z^{-1} \in B$, $z^{-1}$ is integral over $A$, in other words there is an equation of integral dependence

where the $a_i \in A$; now this relation shows that

hence $A$ is certainly a field.

Let us write $S = A - \mathfrak{p}$; then $S^{-1}A'$ is integral over $S^{-1}A$ ($§ 1$, no. 5, Proposition

16), $S^{-1}p$ is a maximal ideal of $S^{-1}A$ (Chapter 11, § 2, no. 5, Proposition 11), $S^{-1}a'$ and $S^{-1}p'$ are ideals of $S^{-1}A'$ lying above $S^{-1}p$ (Lemma 1) and $S^{-1}a' \supseteq S^{-1}p'$. As $S^{-1}p'$ is prime, it is maximal by Proposition 1 and hence $S^{-1}p' = S^{-1}a'$; therefore $a'$ is contained in the saturation of $p'$ with respect to $h(S)$, which is equal to $p'$ (Chapter II, § 2, no. 5, Proposition 11).

If $a'$ is the kernel off, the hypothesis means that $a' \cap A = (0)$; as $A'$ is an integral domain, Corollary 1 may be applied taking $p$ and $p'$ to be the ideal (0) of $A$ and the ideal (0) of $A'$ respectively, whence $a' = (0)$.

For a prime ideal of $A'$ to contain $mA'$, it is necessary and sufficient that its inverse image under $h$ contain $m$ and hence that it lie above $m$, since $m$ is maximal in $A$; the $m'_j$ are therefore the only prime ideals of $A'$ containing $mA'$ (Proposition 1) and therefore $c' = \bigcap_j m'_j$ is the radical of $mA'$ (Chapter II, § 2, no. 6, Corollary 1 to Proposition 13). By definition of $q'_j$, the class mod. $q'_j$ of an element of $A' - m'_j$ is not a divisor of 0 in $A'/q'_j$; on the other hand, as the $m'_j$ are distinct maximal ideals, for every index $j$ there exists an element $a'_j$ belonging to $\bigcap_j m'_j$ and not to $m'_j$ (Chapter II, § 1, no. 1, Proposition 4): then, for all $x \notin m'_j, a'_j x \in r'$, hence the class mod. $q'_j$ of $a'_j x$ is nilpotent and, as that of $a'_j$ is not a divisor of 0, we conclude that the class of $x$ is nilpotent; in other words $m'_j$ is the radical of $q'_j$, which proves (i). It follows that the $q'_j$ are relatively prime in pairs (Chapter II, § 1, no. 1, Proposition 3); (iii) will therefore be a consequence of (ii), taking account of Chapter II, § 1, no. 2, Proposition 5. To establish (ii), we note that in the ring $A'/mA'$ the $m'_j/mA'$ are the only maximal ideals and $q'_j/mA'$ is the saturation of (0) with respect to $m'_j/mA'$ (Chapter 11, § 2, no. 4); we may therefore restrict our attention to the case $mA' = (0)$; the assertion of (ii) then follows from Chapter II, § 3, no. 3, Corollary 2 to Theorem 1.

Remark (1) If $A'$ is Noetherian, it follows from (i) and (ii) that $(q'_j)_{1 \leq j \leq n}$ is the unique primary decomposition of $mA'$ (Chapter IV, § 2, no. 3).

Suppose first that $A$ is a local ring and $p$ the maximal ideal of $A$; then, for every maximal ideal $m'$ of $A'$, $h^{-1}(m')$ is a maximal ideal of $A$ (Proposition 1) and hence equal to $p$, which proves the theorem in this case (since $A'$ contains $A$ by hypothesis and is therefore not reduced to 0). In the general case, let us write $S = A - p$; then $S^{-1}A$ is a local ring whose maximal ideal is $S^{-1}p$ (Chapter 11, § 2, no. 5, Proposition 11), $S^{-1}h : S^{-1}A \to S^{-1}A$, is injective (Chapter II, § 2, no. 4, Theorem 1) and $S^{-1}A'$ is integral over $S^{-1}A$ ($§ 1$, no. 5, Proposition 16); then there exists a prime ideal $q'$ of $S^{-1}A'$ lying above $S^{-1}p$ and we know that $q' = S^{-1}p'$, where $p'$ is a prime ideal of $A'$ lying above $p$ (Lemma 1).

If $h : A \to A'$ is not injective, Theorem 1 is no longer necessarily true, as the example of the homomorphism $Z \to \mathbf{Z}/n\mathbf{Z} \ (n > 1)$ shows. However Theorem 1 can be applied to the canonical injection $h(A) \to A'$; in other words, the statement of Theorem 1 is true for prime ideals $p$ containing $\mathrm{Ker}(h)$.

$pA' \subset p'$ and $\bar{h}(p') = p$.

If $h_1 : A/a \to A'/a'$ is the homomorphism derived from $h$ by taking quotients, $h_1$ is injective by hypothesis and $A'/a'$ is integral over $A/a$ ($§ 1$, no. 1, Proposition 2); then there exists a prime ideal $p'/a'$ of $A'/a'$ (p' prime in $A$) lying above $p/a$ (Theorem 1) and $p'$ is the required ideal.

Let $32$ be the Jacobson radical of $A$. For every maximal ideal $m'$ of $A'$, $m' \cap A$ is a maximal ideal of $A$ (Proposition 1), hence $\mathscr{R} \subset m' \cap A$ and therefore $\mathscr{R} \subset \mathscr{R}' \cap A$ (Algebra, Chapter VIII, § 5, no. 3, Definition 3). Conversely, let $x \in 32' \cap A$; for every maximal ideal $m$ of $A$, there exists a prime ideal of $A'$ lying above $m$ (Theorem 1) and this ideal $m'$ is necessarily maximal (Proposition 1), hence $x \in m' \cap A = m$ and therefore $x \in 32$.

Let $\mathfrak{p}$ be the kernel of $f$, which is a prime ideal since $f(A) \subset L$ is an integral domain; let $\mathfrak{p}'$ be a prime ideal of $A'$ lying above $\mathfrak{p}$ (Theorem 1). Then $A/\mathfrak{p}$ is canonically identified with a subring of $A'/\mathfrak{p}'$ and $A'/\mathfrak{p}'$ is integral over $A/\mathfrak{p}$ ($§ 1$, no. 1, Proposition 2). The homomorphism $f$ defines, by taking the quotient, an isomorphism of $A/\mathfrak{p}$ onto the subring $f(A)$ of $L$, which can be extended to an isomorphism $g$ of the field of fractions $K$ of $A/\mathfrak{p}$ onto a subfield of $L$. As the field of fractions $K'$ of $A'/\mathfrak{p}'$ is algebraic over $K$, $g$ can be extended to an isomorphism $g'$ of $K'$ onto a subfield of $L$ (Algebra, Chapter V, $§ 4$, no. 2, Corollary to Theorem 1); if $\pi': A' \to A'/\mathfrak{p}'$ is the canonical homomorphism, $g' \circ \pi'$ is a homomorphism from $A'$ to $L$ extending $f$.

Remark (2). Let $h : A \to A'$ be a ring homomorphism such that $A'$ is integral over $A$; then the associated continuous mapping $^a h : \mathrm{Spec}(A') \to \mathrm{Spec}(A)$ is closed. For every ideal $a'$ of $A'$, $A'/a'$ is integral over $A'$, hence also over $A$ ($§ 1$, no. 1, Proposition 6) and $\mathrm{Spec}(A'/a')$ is identified with a closed subspace $V(a')$ of $\mathrm{Spec}(A')$; to show that $^a h$ is closed, we see then (replacing $A'$ by $A'/a'$) that it is sufficient to prove that the image of $\mathrm{Spec}(A)'$ under $^a h$ is a closed subset of $\mathrm{Spec}(A)$; now it follows from Theorem 1 that this image is just the set of prime ideals of $A$ containing the ideal $\mathrm{Ker}(h)$ and this set is closed by definition of the topology on $\mathrm{Spec}(A)$.

In fact, by definition $h(S) \subset S'$ and, as

it suffices to prove, by virtue of Chapter II, $§ 2$, no. 3, Proposition 8, that, if a prime ideal $q'$ of $A'$ does not meet $h(S)$, it does not meet $S'$ either. Now, suppose that $q' \cap h(S) = \varnothing$ and let $q = h^{-1}(q')$; then $q \cap S = \varnothing$, in other words $q \subset \mathfrak{p}$. As $q'$ lies above $q$ by definition, it follows from Corollary 2 to Proposition 1 that there is an index $c$ such that $q' \subset p'_c$ and hence $q' \cap S' = \varnothing$, which completes the proof.

Let $S = A - p$; by Lemma 1 we may replace $A$ by $S^{-1}A$, $A'$ by $S^{-1}A'$ (which is a finitely generated $S^{-1}A$-module) and $p$ by $S^{-1}p$; in other words, we may assume that $A$ is a local ring and $p$ is its maximal ideal. Then (by the remark made at the beginning of this no.) $A$ may be replaced by $A/p$, $A'$ by $A'/pA'$ and $p$ by $(0)$, for $A'/pA' = (A/p) \otimes_A A'$ is a finitely generated $(A/p)$-module. Thus we have finally reduced the problem to proving the proposition when $A$ is a field and $p = (0)$; $A'$ is then an $A$-algebra of finite rank and therefore Artinian and we know that in such an algebra there is only a finite number of prime ideals (Chapter IV, § 2, no. 5, Proposition 9).

As the elements of $S$ are invariant under $\mathscr{G}$, clearly, if $\sigma.p' = p'$, also $\sigma.(S^{-1}p') = S^{-1}p'$. Conversely, suppose that $\sigma \in \mathscr{G}$ is such that $\sigma.(S^{-1}p') = S^{-1}p'$; then, if $x \in p'$, $(\sigma.x)/1 \in S^{-1}p'$ and there therefore exists $s \in S$ such that $s(\sigma.x) \in p'$, whence $\sigma.x \in p'$ since $p'$ is prime and $s \notin p'$; this proves that $\sigma.p' \subset p'$ and it can be similarly shown that $\sigma^{-1}.p' \subset p'$, hence $\sigma.p' = p'$ and $\sigma \in \mathscr{G}^{\mathbf{Z}}(p')$. If $\sigma \in \mathscr{G}^{\Gamma}(p')$, then $\sigma.x - x \in p'$ for all $x \in A'$, hence also, for all $s \in S$,

and therefore $\sigma \in \mathscr{G}^{\Gamma}(S^{-1}p')$. Conversely, suppose that $\sigma \in \mathscr{G}^{\Gamma}(S^{-1}p')$; then, for all $x \in A'$, $\sigma.(x/1) - (x/1) \in S^{-1}p'$ and therefore there exists $s \in S$ such that $s(\sigma.x - x) \in p'$, whence as above $\sigma.x - x \in p'$, which proves that $\sigma \in \mathscr{G}^{\Gamma}(p')$. The last two assertions follow from § 1, no. 9, Proposition 23.

(i) Given two prime ideals $p', q'$ of $A'$ lying above the same prime ideal $p$ of $A$, there exists $\sigma \in \mathscr{G}$ such that $q' = \sigma.p'$; in other words, $\mathscr{G}$ operates transitively on the set of prime ideals of $A'$ lying above $p$.

(ii) Let $p'$ be a prime ideal of $A'$, $p = p' \cap A$ and $k$ (resp. $k'$) the field of fractions of $A/p$ (resp. $A'/p'$). Then $k'$ is a quasi-Galois extension () of $k$ and the canonical homomorphism $\sigma \mapsto \bar{\sigma}$ from $\mathscr{G}^{\mathbf{Z}}(p')$ to the group $\Gamma$ of $k$-automorphisms of $k'$ defines, by taking the quotient, an isomorphism of $\mathscr{G}^{\mathbf{Z}}(p')/\mathscr{G}^{\Gamma}(p')$ onto $\Gamma$.*

(*) In order to avoid confusion with other meanings of the word "normal" we henceforth use the term "quasi-Galois extension" as synonymous with the term "normal extension" defined in Algebra, Chapter V, § 6, no. 2, Definition 2.

(i) If $x \in \mathfrak{q}$ then $\prod_{\sigma \in \mathscr{G}} \sigma.x \in q' \cap A = p \subset p'$; then there exists $\sigma \in \mathscr{G}$ such that $\sigma.x \in p'$, that is $x \in \sigma^{-1}.q'$. We conclude that $q' \subset \bigcup_{\sigma \in \mathscr{G}} \mathfrak{s}.\mathfrak{p}'$ and hence (since $\mathscr{G}$ is finite and the $\mathfrak{s}.\mathfrak{p}'$ prime) there exists $\sigma \in \mathscr{G}$ such that $\mathfrak{q}' \subset \mathfrak{s}.\mathfrak{p}'$ (Chapter II, § 1, no. 1, Proposition 2); as $q'$ and $\sigma.\mathfrak{p}'$ both lie above $p$, $q' = \sigma.\mathfrak{p}'$ (no. 1, Corollary 1 to Proposition 1).

(ii) To see that $k'$ is a quasi-Galois extension of $k$, it suffices to prove that every element $\bar{x} \in A'/p'$ is a root of a polynomial $P$ in $k[X]$ all of whose roots are in $A'/p'$ (Algebra, Chapter V, § 6, no. 3, Corollary 3 to Proposition 9). Now, let $x \in A'$ be a representative of the class $\bar{x}$; the polynomial $Q(X) = \prod_{\sigma \in \mathscr{G}} (X - \sigma.x)$ has all its coefficients in $A$; let $P(X)$ be the polynomial in $(A/p)[X]$ whose coefficients are the images of those of $Q$ under the canonical homomorphism $\pi : A \to A/p$. As $\pi$ may be considered as the restriction to $A$ of the canonical homomorphism $\pi' : A' \to A'/p'$, it is seen that $P$ is the product in $(A'/p')[X]$ of the linear factors $X - \pi'(\sigma.x)$ and therefore solves the problem since $\bar{x} = \pi'(x)$.

Clearly, for all $\sigma \in \mathscr{G}^z$, $\hat{\sigma}$ is a $k$-automorphism of $k'$; it remains to verify that $\sigma \mapsto \hat{\sigma}$ maps $\mathscr{G}^z$ onto the group of all $k$-automorphisms of $k'$. Let us write $S = A - p$; $k$ and $k'$ are not changed by replacing $A'$ and $p'$ by $S^{-1}A'$ and $S^{-1}p'$ respectively, by virtue of § 1, no. 9, Proposition 23 and the relation $S^{-1}p' \cap S^{-1}A = S^{-1}(A \cap p') = S^{-1}p$ (Chapter 11, § 2, no. 4); it follows from Lemma 3 that neither $\mathscr{G}^z$ nor its operation on $k'$ is changed; we may therefore restrict our attention to the case where $p$ is maximal, in which case we know that so is $p'$ (no. 1, Proposition 1) and every element of $k'$ is therefore of the form $\pi'(x)$ for some $x \in A'$; it has been seen above that such an element is a root of a polynomial in $k[X]$ of degree $\leq \mathrm{Card}(\mathscr{G})$. As every finite separable extension of $k$ admits a primitive element (Algebra, Chapter V, § 7, no. 7, Proposition 12 and § 11, no. 4, Proposition 4), it is seen that every finite separable extension of $k$ contained in $k'$ is of degree $\leq \mathrm{Card}(\mathscr{G})$, whence it follows that the greatest separable extension $k'_s$ of $k$ contained in $k'$ (Algebra, Chapter V, § 7, no. 6, Proposition 11) is of degree $\leq \mathrm{Card}(\mathscr{G})$ (Algebra, Chapter V, § 3, no. 2, Remark 2). Let $y \in A'$ be an element such that $\pi'(y)$ is a primitive element of $k'_s$. The ideals $\sigma.p'$ for $\sigma \in \mathscr{G} - \mathscr{G}^z$ are maximal and distinct from $p'$ by definition; there therefore exists $x \in A'$ such that $x \equiv y \pmod{p'}$ and $x \in \sigma^{-1}p'$ for $\sigma \in \mathscr{G} - \mathscr{G}^z$ (Chapter 11, § 1, no. 2, Proposition 5). This being so, let $u$ be a $k$-automorphism of $k'$ and let $P(X) = \prod_{\sigma \in \mathscr{G}} (X - \pi'(\sigma.x))$; as $\pi'(x)$ is a root of $P$ and $P \in k[X]$, $u(\pi'(x))$ is also a root of $P$ in $k'$ and hence there exists $\tau \in \mathscr{G}$ such that

but $u(\pi'(x)) \neq 0$ and, for $\sigma \in \mathscr{G} - \mathscr{G}^z$, $\sigma.x \in p'$ and hence $\pi'(\sigma.x) = 0$; we conclude that necessarily $\tau \in \mathscr{G}^z$. But as $u$ and $\tilde{\tau}$ have the same value for the primitive element $\pi'(y) = \pi'(x)$ of $k'_s$, they coincide on $k'_s$ and, as $k'$ is a radicial extension of $k'_s$, they coincide on $k'$.