Valuations on tensor products Let $A$ be a commutative ring, $B$ (resp. $C$) be a commutative $A$-algebra endowed with a valuation $v$ (resp. $w$), not necessarily of rank 1. Assume that $v$ and $w$ induce equivalent valuations on $A$. How to construct a valuation $u$ on $B\otimes_A C$ extending $v$ and $w$?
Without loss of generality, we may assume $A$, $B$, $C$ to be fields. If $B$ is an algebraic extension of $A$, the existence of $u$ follows from the fact that extensions of a valuation to a normal extension field are conjugate to each other [Bourbaki, AC VI 8 Prop. 7]. Thus the only case left to check is when both $B$ and $C$ are purely transcendental over $A$.
Huber lists the existence of $u$ as a "simple property" of valuations [Etale cohomology of Rigid Analytic Varieties and Adic Spaces, 1.1.14 f].  No proof is given there. Are there other references for this?
Added on Aug. 5: Let us denote the value groups of $A$, $B$, $C$ by $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, respectively. The value group of $u$ is an extension of $\Gamma_B$ and $\Gamma_C$ over $\Gamma_A$. How to construct such an extension of linearly ordered Abelian groups? We could put the lexicographic order on $\Gamma_B\times \Gamma_C$, but then we cannot quotient out by the diagonal image of $\Gamma_A$ as the image is not convex.
 A: I spent a while looking for a proof in the literature and haven't found one yet. Fortunately, this is easily resolved using a bit of model theory. (WARNING: I am not a model theorist!)
By Zorn's lemma, we may also assume that $C = A(x)$. By the case already discussed, we may also assume that $A$ and $B$ are algebraically closed; it is also harmless to assume that $A$ has nontrivial valuation.
Let ACVF be the theory of algebraically closed (nontrivially) valued fields. This theory is model-complete (see Completeness of Algebraically Closed Valued Fields(ACVF) Theory); consequently, for any finite set of polynomials $P_1, \dots, P_n$ over $A$, one can find an extension $D$ of $B$ and an element $y$ of $D$ such that for each $i$, $P_i(x)$ has valuation $\geq 0$ iff $P_i(y)$ does. Using the compactness of the Riemann-Zariski space of $B$ for the constructible topology, we can then construct a valuation on $B(x)$ that has the correct value on every element of $A(x)$, as needed.
A: In terms of valuation rings the question is equivalent to the following: given valuation rings $A, B, C$ and injective local ring homomorphisms $A \to B$ and $A \to C$ there exists a ring map $B \otimes_A C \to D$ where $D$ is a valuation ring such that $B \to D$ and $C \to D$ are injective local ring homomorphisms.
To prove this, it suffices to find a specialization $x' \leadsto x$ of points of $\text{Spec}(B \otimes_A C)$ such that $x'$ maps to the generic points of $\text{Spec}(B)$ and $\text{Spec}(C)$ and such that $x$ maps to the closed points of $\text{Spec}(B)$ and $\text{Spec}(C)$. Namely, then we can apply Tag 01J8 to find $D$.
Denote $\kappa_A$ the residue field of $A$ and similarly for $B$ and $C$. Since $\kappa_B \otimes_{\kappa_A} \kappa_C$ is not the zero ring, there exists a point $x$ of $\text{Spec}(B \otimes_A C)$ mapping to the closed points of $\text{Spec}(B)$ and $\text{Spec}(C)$. Pick any maximal point $x'$ of $\text{Spec}(B \otimes_A C)$ specializing to $x$, in other words, $x'$ corresponds to a minimal prime ideal of $B \otimes_A C$. Since $A \to B$ and $A \to C$ are flat ring maps (as torsion free $A$-modules are flat), the ring maps $B \to B \otimes_A C$ and $C \to B \otimes_A C$ are flat as well. By going down for flat ring maps, we see that $x'$ maps to the generic points of $\text{Spec}(B)$ and $\text{Spec}(C)$. This finishes the proof.
