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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.

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Isn't this just a matter of saying $v(b \otimes c) = v(b) + v(c)$ and letting the valuation of an element of $B \otimes C$ be the sup, over all possible ways to write the element as $\sum b_i \otimes c_i$, of $\inf_i v(b_i \otimes c_i)$? –  Laurent Berger Aug 4 '12 at 16:32
    
I don't see how this works for valuations of higher rank, as the inf might not exist. –  Weizhe Zheng Aug 5 '12 at 2:08

1 Answer 1

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.

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