A valuation overring has a center

Let $\mathfrak{m}$ be the set of nonunits of $V$. If $a$ is a nonunit and $v\in V$, then $va$ cannot be a unit, since an inverse of $va$ would give an inverse of $a$. If nonunits $a,b$ have nonzero sum, [F1] gives either $a\mathrm{/}b\in V$ or $b\mathrm{/}a\in V$ when both are nonzero; in the first case $a+b=b(1+a\mathrm{/}b)$ is a nonunit, and similarly in the second. The zero cases are immediate. Thus $\mathfrak{m}$ is a proper ideal, and every element outside it is a unit. It is consequently the unique maximal ideal; it is prime because a product that is a unit has unit factors.

Its inverse image $\mathfrak{q}$ in $D$ is prime. Every element of $D\setminus \mathfrak{q}$ maps to a unit of $V$, so the inclusion extends by fractions to $D_{\mathfrak{q}}\to V$. It is injective because both rings embed in $K$. A fraction $a\mathrm{/}s$ with $s\mathrm{\notin }\mathfrak{q}$ maps into $\mathfrak{m}$ exactly when $a\in \mathfrak{q}$, proving locality.