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.

  • $\begingroup$ 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 $\inf_i v(b_i \otimes c_i)$? $\endgroup$ Aug 4 '12 at 16:32
  • 2
    $\begingroup$ I don't see how this works for valuations of higher rank, as the inf might not exist. $\endgroup$ Aug 5 '12 at 2:08
  • $\begingroup$ Isn't it the content of Exercise 2 of Bourbaki, AC VI (Valuations), §2 (page 167), which, however, is stated in the language of “places”? $\endgroup$
    – ACL
    Apr 3 '15 at 20:26
  • $\begingroup$ The construction asked in the addendum of this question seems to be equivalent to finding a total order on $\Gamma_B\times\Gamma_C$, such that it restricts to the given total order on $\Gamma_B$ and $\Gamma_C$ and asking the intersection of $\Gamma_A$(diagonal) with convex hull (under the total order we found) of anti-diagonal $\Gamma_A$ is identity. Such an order can't be lexicographic order as one can check easily, and it's certainly not obvious that it actually exists. $\endgroup$
    – S. Li
    Aug 7 '16 at 22:03

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.

  • 3
    $\begingroup$ The possibility of the construction ask for by Weizhe Zheng is called amalgamation property (AP) by model theorists. They observed its importance in mathematics hence studied its consequences in general algebraic structures such as fields, ordered groups, valued fields... Since ACVF plays a large rôle in applied model theorists, there should not be a surprise that these mathematicians know such facts. As shown in model theory, AP follows from the fact that the model-companion satisfies quantifier elimination. In the present case, the model-companion is ACVF and satisfies QE by Robinson. $\endgroup$
    – ACL
    Apr 3 '15 at 20:33

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.


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