If $m(A) \subset m(B)$, $m(B) \cap A$ is an ideal of $A$ which does not contain 1 and which contains the maximal ideal $m(A)$; it is therefore equal to it and (a) implies (b). If $B$ dominates $A$, the ideal $Bm(A)$ is contained in $m(B)$ and hence does not contain 1; thus (b) implies (c). If (c) holds, $Bm(A)$ is contained in the unique maximal ideal $m(B)$ of $B$, whence (a).

Note that, if $K$ is a ring, the relation "B dominates A" is an order relation on the set of local subrings of $K$.

Let $A$ and $B$ be two local rings such that $B$ dominates $A$. The canonical injection $A \to B$ defines by taking quotients an isomorphism of the field $\kappa(A)$ onto a subfield of $\kappa(B)$; this isomorphism allows us to identify $\kappa(A)$ with a subfield of $\kappa(B)$.

Examples (1) Let $A$ be a Noetherian local ring and $\hat{A}$ its completion; the local ring $\hat{A}$ then dominates $A$ (Chapter 111, § 3, no. 5, Proposition 9). (2) Let $B$ be an integral domain, $A$ a subring of $B$, $p'$ a prime ideal of $B$ and $p = A \cap p'$. Then $pA_p \subset p'B_{p'}$, so that $B_{p'}$ dominates $A_p$.

We shall show the theorem by proving the following implications

Suppose that (a) holds. Then $V$ is a local ring. Let $L$ be an algebraic closure of the residue field $\kappa(V)$ and $h$ the canonical homomorphism from $V$ to $L$. Let $V'$ be a subring of $K$ containing $V$ and $h'$ a homomorphism from $V'$ to $L$ extending $h$. If $p'$ is the kernel of $h'$, then $p' \cap V = m(V)$; hence (no. 1, Example 2) $V'_{p'}$ dominates $V$, which implies $V'_{p'} = V$ and $V' = V$. Thus (b) holds.

Suppose (b) holds. Let $L$ be an algebraically closed field and $h$ a homomorphism from $V$ to $L$; suppose that $h$ is maximal in the set of homomorphisms from subrings of $K$ to $L$; let $p$ be the kernel of $h$. The elements of $h(V - p)$ being invertible in $L$, $h$ can be extended to a homomorphism from $V_p$ to $L$ (Chapter II, § 2, no. 1, Proposition 1); hence $V = V_p$, which shows that $V$ is a local ring and that $p$ is its maximal ideal. Let $x$ be a non-zero element of $K$; we must show that one at least of the elements $x, x^{-1}$ belongs to $V$, that is, by virtue of the maximal character of $h$, that $h$ can be extended to $V[x]$ or $V[x^{-1}]$. If $x$ is integral over $V$, this follows from Chapter V, § 2, no. 1, Corollary 4 to Theorem 1. If $x$ is not integral over $V$, we shall use the following lemma:

As $x$ is not integral over $A$, it is not nilpotent and therefore $B, \neq 0$; moreover, $x/1 \notin A[1/x]$, otherwise there would be a relation of the form

for some $n \geq 0$ (where $a_i \in A$ for $0 \leq i \leq n$), which is equivalent to

for some convenient $h \geq 1$; but such a relation would imply that $x$ is integral over $A$, contrary to the hypothesis. The existence of a maximal ideal of $A[1/x]$ containing $1/x$ therefore follows from the fact that $1/x$ is not invertible in $A[1/x]$ (Algebra, Chapter I, § 8, no. 7, Theorem 2).

Then let $\mathfrak{M}$ be a maximal ideal of $A[1/x]$ containing $1/x$; let $\phi : A \to A[1/x]$, $p : A[1/x] \to A[1/x]/\mathfrak{M}$ be the canonical homomorphisms; then

since $p(1/x) = 0$; this proves that $p(\phi(A))$ is a field and hence the inverse image $\phi^{-1}(\mathfrak{M})$ is a maximal ideal of $A$.

We apply this lemma with $A = V$ and $B = K$; then there is a maximal ideal $\mathfrak{m}$ of $V[x^{-1}]$ containing $x^{-1}$ and $\mathfrak{m} \cap V$ is a maximal ideal of $V$, then $\mathfrak{m} \cap V = p$ since $V$ is local; denoting by $f$ the canonical homomorphism of $V[x^{-1}]$ onto $V[x^{-1}]/\mathfrak{m}$, $f(x^{-1}) = 0$, whence $V/p = f(V) = f(V[x^{-1}])$; as $h$ defines by taking the quotient an injective homomorphism $\bar{h}$ from $V/p$ to $L$, $\bar{h}$ of is a homomorphism from $V[x^{-1}]$ to $L$ extending $h$. Hence (c) holds.

Suppose now that (c) holds. Clearly $K$ is the field of fractions of $V$. Let $a$ and $b$ be elements of $V$ such that $Va \notin Vb$; we show that $Vb \subset Va$. It is true if $b = 0$; otherwise the relation $a \notin Vb$ implies $b^{-1}a \notin V$, whence by (c) $a^{-1}b \in V$ and therefore $Vb \subset Va$. Hence (d) holds.

Suppose that (d) holds. Let $a$ and $b$ be ideals of $V$ such that $a \notin b$. There exists $a \in a$ such that $a \notin b$. For all $b \in b$, $a \notin Vb$, whence $Va \notin Vb$ and therefore $Vb \subset Va \subset a$ (by (c)) and $b \in a$. Then $b \subset a$, which shows that condition (e) is satisfied.

Suppose finally that (e) holds. As $V$ has a maximal ideal, it has only one and is therefore a local ring. Let $V'$ be a local subring of $K$ dominating $V$ and let $x$ be a non-zero element of $V'$, we write $x = ab^{-1}$ where $a \in V, b \in V$. One of the ideals $Va, Vb$ is contained in the other. If Va \subset Vb, then $x \in V$. If Vb \subset Va, then $x^{-1} \in V$; as the ideal $V' m(V)$ does not contain 1 (no. 1, Proposition 1), $x^{-1} \notin m(V)$, whence again $x \in V$ since $V$ is local. Every element of $V'$ therefore belongs to $V$; we conclude that (a) holds.

Let $\mathfrak{S}$ be the set of homomorphisms of subrings of $K$ to $L$, ordered by the relation of extension. This set is inductive; for if $(h_\alpha)_{\alpha \in I}$ is a non-empty totally ordered family of elements of $\mathfrak{S}$ and $B$, is the defining ring of $h_\alpha$, the $B_\alpha$ form a totally ordered family of subrings of $K$ and their union $B$ is therefore a subring of $K$; there therefore exists a unique mapping $\bar{h}$ from $B$ to $L$ which extends the $h_\alpha$ (Set Theory, Chapter II, § 4, no. 6, Proposition 7) and it is immediately seen that $\bar{h}$ is a homomorphism from $B$ to $L$. Zorn's Lemma then shows that there exists a maximal element $h'$ of $\mathfrak{S}$ which extends $h$. The defining ring $V$ of $h'$ is a valuation ring of $K$ (Theorem 1); if $p$ is the kernel of $h'$, $h'$ can be extended to a homomorphism from $V_p$ to $L$ (Chapter II, § 2, no. 1, Proposition 1), whence $V_p = V$ and $p = m(V)$.

Apply Theorem 2 to the canonical homomorphism $h$ from $A$ to an algebraic closure $L$ of $A/m(A)$.

Let $x$ be an element of $A'$ and $V$ a valuation ring of $K$ containing $A$; as $x$ is integral over $V$, there exists a prime ideal $p'$ of $V[x]$ such that $p' \cap V = m(V)$ (Chapter V, § 2, no. 1, Theorem 1); clearly then the local ring $(V[x])_{p'}$ dominates $V$ and hence is equal to it; whence $x \in V$. Conversely let $y$ be an element of $K$ which is not integral over $A$; then there exists a maximal ideal 2.2 of $A[y^{-1}]$ which contains $y^{-1}$ (no. 2, Lemma 1); there exists also a valuation ring $V$ of $K$ which dominates $(A[y^{-1}])_m$ (no. 2, Corollary to Theorem 2); as $y^{-1} \in m(V)$, $y \notin V$. Further $2.2 \cap A$ is a maximal ideal of $A$ (no. 2, Lemma 1); hence, if $A$ is local, $2.2 \cap A = m(A)$ and $V$ dominates $A$.

In the case of a Noetherian ring, Corollary 2 can be made more precise (Chapter VII, § 1, no. 3, Corollary to Theorem 1).

Theorem 1 (c) show that, if V' is a valuation ring of K', V' $\cap K$ is a valuation ring of K and V' dominates V' $\cap K$. For V' to dominate A, it is necessary and sufficient that V' $\cap K$ dominate A and therefore be equal to it.

Let A be filtered by the Ap^n and let v denote the corresponding order function (Chapter III, § 2, no. 2). As

the relation $v(x) = +\infty$ implies $x = 0$. Let a be an ideal $\neq (0)$ of A and a an element of a at which v takes its least value; let us write $v(a) = s$ ($s \neq +\infty$). Then $a \subset Ap^s$. In particular, there exists $u \in A$ such that $a = up^s$; as $a \notin Ap^{s+1}$, $u \notin Ap$; hence u is invertible and $p^s \in A$ $a \subset a$. It follows that $a = Ap^s$, whence our first assertion. It is also seen that every element $a \neq 0$ of A may be written in the form $a = up^{v(a)}$ where u is invertible. If $a' = u'p^{v(a')}$ ($u'$ invertible) is another non-zero element of A, then $aa' = uu'p^{v(a)+v(a')}$; hence, if p is not nilpotent, $aa' \neq 0$ and A is an integral domain. Then, as the set of ideals of A is totally ordered by inclusion, we conclude that A is a valuation ring (Theorem I (e)).

For example, if $p$ is a prime number, the local ring $\mathbf{Z}_{(p)}$ is a valuation ring. Let $B = K[X_1, \ldots, X_n]$ be the polynomial ring in $n$ indeterminates over a field $K$; the ideal $BX_1$ is prime, since $B/BX_1$ is isomorphic to $K[X_2, \ldots, X_n]$; hence $B_{BX_1}$ is a valuation ring; it is composed of rational functions $PQ^{-1}$, where $P$ and $Q$ are polynomials and $Q(0, X_2, \ldots, X_n) \neq 0$.

More generally, we shall see that, if $F$ is an extremal element of

$B_{BF}$ is a valuation ring (cf. Chapter VII, § 3, no. 5). \*

The ring of formal power series $K[[T]]$ in one indeterminate over a field $K$ is a Noetherian local integral domain whose maximal ideal is principal; it is therefore a valuation ring. On the other hand the ring $K[[T_1, T_2]]$ of formal power series in two indeterminates, which is a Noetherian local integral domain is not a valuation ring, for neither of the elements $T_1, T_2$ is a multiple of the other.

Clearly $A_{Ap}$ ($p$ extremal) is a valuation ring containing $A$ and distinct from $K$ (Proposition 2). Conversely, let $V$ be a valuation ring distinct from $K$ and containing $A$. As $V \neq K$, $m(V)$ contains an element $x \neq 0$; writing $x = a/b$ where $a \in A$ and $b \in A$, it is seen that $A \cap m(V)$ contains the non-zero element $a$. As $A \cap m(V)$ is prime, it is of the form $Ap$ where $p$ is extremal in $A$. Then $A_{Ap} \subset V, pA_{Ap} \subset m(V)$, so that $V$ dominates $A_{Ap}$; as $A_{Ap}$ is a valuation ring of $K$ (Proposition 2), $V = A_{Ap}$.

Every subring of $\mathbf{Q}$ contains $\mathbf{Z}$.

If $X \in V$, then $K[X] \subset V$ and the assertion made follows from Proposition 3. If $X \notin V$, then $X^{-1} \in V$, whence $K[X^{-1}] \subset V$; then $V$ is the local ring of $K[X^{-1}]$ at a prime ideal $p$ (Proposition 3) and this ideal contains $X^{-1}$ since $X^{-1}$ is not invertible in V; then $p = X^{-1}K[X^{-1}]$ since the latter ideal is maximal. Finally let us consider a rational function $A(X)/B(X)$, where A and B are polynomials of respective degrees $a$ and $b$; then $A(X) = X^a A'(X^{-1})$ and $B(X) = X^b B'(X^{-1})$, where $A'$ and $B'$ are polynomials such that $A'(0) \neq 0$ and $B'(0) \neq 0$; hence, for $A(X)/B(X)$ to belong to the local ring of $K[X^{-1}]$ at $X^{-1}K[X^{-1}]$, it is necessary and sufficient that $a \leq b$.