Lemma 1

Let $A$ be a subring of a ring $B$ and $x$ an element of $B$ not integral over $A$; then the ring of fractions $B$, (Chapter II, § 5, no. 1) is not reduced to O and there exists in the subring $A[1/x]$ of $B$, maximal ideals containing $1/x$; moreover, if $\mathfrak{M}$ is any of these maximal ideals, the inverse image of $\mathfrak{M}$ in $A$ is a maximal ideal.

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.