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.