Suppose $(R, \mathfrak m)$ is a local domain, and $R_v \subset Q(R)$ is a valuation ring with maximal ideal $\mathfrak m_v$, which dominates $R$. This means we have $R \subset R_v$ and $\mathfrak m = \mathfrak m_v \cap R$. Then there is a canonical field extension
$$R / \mathfrak m \hookrightarrow R_v / \mathfrak m_v.$$
I wonder if this is an isomorphism? If not, what is a counterexample?
For example if $R$ is the local ring of a closed point in a variety over an algebraically closed field $k$, this is always true because $R / \mathfrak m \cong k$.
For a bit of context, I'm reading a paper by Mumford [1] where he considers two varieties $Z \subset V$, and has a valuation ring $R_1 \subset k(Z)$, and a valuation ring $R_2 \subset k(V)$ such that $R_2$ dominates the local ring of $Z$, $\mathcal O_{V, Z}$. Mumford then wants to take the composite of $R_1$ and $R_2$, which is usually the preimage of $R_1$ under the projection $R_2 \to R_2 / \mathfrak m_2$. However, I think to call this the composite we need $k(Z) = R_2 / \mathfrak m_2$, so that $R_1$ has fraction field $R_2/\mathfrak m_2$. At least this is the situation in which Matsumura [2, Ch. 10] defines the composite of valuation rings.
[1] Mumford, David; An analytic construction of degenerating abelian varieties over complete rings
[2] Matsumura, Hideyuki; Commutative Ring Theory
Best Answer
Here is a counter-example. Consider the blow-up $$X = \operatorname{Bl}_0 \mathbb A^2 \to \mathbb A^2,$$ and let $E \subset X$ be the exeptional curve. According to [1, Exc. II.4.12 (b.2)], the local ring $R_v = \mathcal O_{X, \varepsilon}$ at the generic point $\varepsilon \in E$ is a valuation ring which dominates the local ring $R = \mathcal O_{\mathbb A^2, 0}$. Let's calculate this! We have $$X = \{(x,y, [\xi:\eta]) \,|\, x \eta = y \xi\} \subset \mathbb A^2 \times \mathbb P^1,$$ so the open set $\{\eta \neq 0\}$ is affine with coordinate ring $\mathbb C[y, x/y] \subset \mathbb C(x,y)$. Localizing at the prime ideal $\mathfrak p = (y)$ yields $$\mathcal O_{X, \varepsilon} = \mathbb C[y, x/y]_{(y)},$$ so we see that the residue field of $\mathcal O_{X, \varepsilon}$ is the function field $\mathbb C(x/y)$, which has transcendence degree $1$ over $\mathbb C$. On the other hand, the residue field of $$\mathcal O_{\mathbb A^2, 0} = \mathbb C[x,y]_{(x,y)}$$ is just $\mathbb C$, so the two residue fields can never be isomorphic.
For completeness, it is clear that $\mathcal O_{X, \varepsilon}$ is a DVR, because it is a noetherian one-dimensional regular domain. To see that $\mathcal O_{X, \varepsilon}$ dominates $\mathcal O_{\mathbb A^2, 0}$, note that $x = y \cdot \frac{x}{y} \in y \cdot \mathbb C[y, x/y]$. So the inclusion $\mathbb C[x,y]_{(x,y)} \subset \mathbb C[y, x/y]_{(y)}$ maps the maximal ideal $(x,y) \mathcal O_{\mathbb A^2, 0}$ to $y \cdot \mathcal O_{X, \varepsilon}$, and hence $$\mathcal O_{\mathbb A^2, 0} \cap y \cdot \mathcal O_{X, \varepsilon} = (x,y) \mathcal O_{\mathbb A^2, 0}.$$
[1] Hartshorne, Algebraic Geometry