Let
\(\sharp\colon V^\flat \to V\) be the multiplicative map obtained by composing the constant lift
\([\bullet]: V^\flat \to W(V^\flat)\) with the quotient map
\(W(V^\flat) \to V\text{.}\) It is customary to write the image of
\(x\) under
\(\sharp\) as
\(x^\sharp\) rather than
\(\sharp(x)\text{.}\)
Suppose that
\(V\) is a valuation ring. If
\(\Frac V\) has characteristic
\(p\text{,}\) then
\(V = V^\flat\) and there is nothing more to check; we may thus assume that
\(\Frac V\) has characteristic 0. Since
\(V\) is an integral domain, the
\(p\)-power map on
\(V\) is injective; hence for
\(x \in V^\flat\text{,}\) \(x^\sharp = 0\) if and only if
\(x = 0\text{.}\) This in turn implies that
\(V^\flat\) is an integral domain (if
\(xy = 0\) then
\(x^\sharp y^\sharp = 0\)) and that the principal ideals of
\(V^\flat\) are totally ordered with respect to inclusion (if
\(x^\sharp\) is divisible by
\(y^\sharp\text{,}\) then the ratio admits a coherent sequence of
\(p\)-power roots and so is itself in the image of
\(\sharp\)). Hence
\(V^\flat\) is a valuation ring.
Conversely, suppose that
\(V^\flat\) is a valuation ring. Again, we may assume that
\(p \neq 0\) in
\(V\text{;}\) since
\(V^\flat\) is an integral domain, we may apply
Exercise 8.5.3 to deduce that
\(V\) is
\(p\)-torsion-free. Since
\(V^\flat\) is a local ring, so are
\(W(V^\flat)\) and its quotient
\(V\text{.}\) Choose
\(\varpi \in V\) as per
Lemma 8.2.3.
We need to show that given any two nonzero elements
\(x,y \in V\text{,}\) one is a multiple of the other. By dividing by powers of
\(\varpi\) as needed, we may reduce to the case where
\(x\) and
\(y\) have nonzero images in
\(V/\varpi\) and hence in
\(V/p = V^\flat/d\text{.}\) Since
\(V^\flat\) is a valuation ring, after possibly swapping terms we can write
\(x = yz + pu\) for some
\(z,u \in V\text{.}\) Similarly, we can write
\(\varpi \equiv yw + pv\) for some
\(w,v \in V\text{.}\) Since
\(V\) is classically
\(\varpi\)-complete,
\(1 - (p/\varpi)v\) is a unit in
\(V\text{;}\) hence
\(\varpi\) is divisible by
\(y\text{,}\) as then is
\(p\text{.}\) Consequently,
\(x = yz + pu\) is also divisible by
\(y\text{,}\) as desired.