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Algebraically closed fields with several valuation rings Will Johnson March 4, 2018 Will Johnson Multi-valued fiels March 4, 2018 1 / 23 Main results Theorem Let O 1 , . . . , O n be arbitrary valuation rings on K = K alg . The structure ( K

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Algebraically closed fields with several valuation rings

Will Johnson March 4, 2018

Will Johnson Multi-valued fiels March 4, 2018 1 / 23

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Main results

Theorem

Let O1, . . . , On be arbitrary valuation rings on K = K alg. The structure (K, O1, . . . , On) is. . .

1 . . . always NTP2 2 . . . NIP only when the Oi are pairwise comparable. Will Johnson Multi-valued fiels March 4, 2018 2 / 23

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Main results

Theorem

Let O1, . . . , On be arbitrary valuation rings on K = K alg. The structure (K, O1, . . . , On) is. . .

1 . . . always NTP2 2 . . . NIP only when the Oi are pairwise comparable.

Theorem

The (incomplete) theory of n-multi-valued algebraically closed fields is decidable.

Will Johnson Multi-valued fiels March 4, 2018 2 / 23

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Main results

Theorem

Let O1, . . . , On be arbitrary valuation rings on K = K alg. The structure (K, O1, . . . , On) is. . .

1 . . . always NTP2 2 . . . NIP only when the Oi are pairwise comparable.

Theorem

The (incomplete) theory of n-multi-valued algebraically closed fields is decidable. These results are preliminary, though the case of independent valuations is in my dissertation.

Will Johnson Multi-valued fiels March 4, 2018 2 / 23

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E.c. multi-valued fields

Theorem

Consider an n-multi-valued field (K, O1, . . . , On). The following are equivalent: K is existentially closed among n-multi-valued fields. K = K alg, each Oi is non-trivial (Oi = K), and OiOj = K for i = j.

Will Johnson Multi-valued fiels March 4, 2018 3 / 23

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E.c. multi-valued fields

Theorem

Consider an n-multi-valued field (K, O1, . . . , On). The following are equivalent: K is existentially closed among n-multi-valued fields. K = K alg, each Oi is non-trivial (Oi = K), and OiOj = K for i = j. So the model companion of the theory of fields with n valuations. is the theory of algebraically closed fields with n pairwise-independent non-trivial valuations.

Will Johnson Multi-valued fiels March 4, 2018 3 / 23

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Independent topologies

Definition

A collection T1, . . . , Tn of topologies on a set X are independent if U1 ∩ · · · Un = ∅ whenever Ui is a non-empty Ti-open. Equivalently, the diagonal embedding X ֒ →

n

  • i=1

(X, Ti) has dense image.

Will Johnson Multi-valued fiels March 4, 2018 4 / 23

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Independent topologies

Definition

A collection T1, . . . , Tn of topologies on a set X are independent if U1 ∩ · · · Un = ∅ whenever Ui is a non-empty Ti-open. Equivalently, the diagonal embedding X ֒ →

n

  • i=1

(X, Ti) has dense image.

Theorem (Stone approximation)

If T1, . . . , Tn are distinct “valuation-type” topologies on a field K, they are automatically independent.

Will Johnson Multi-valued fiels March 4, 2018 4 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed (b) For any irreducible variety V /K, the valuation topologies on V (K) are independent.

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed (b) For any irreducible variety V /K, the valuation topologies on V (K) are independent. (c) The valuation topologies on A1(K) = K 1 are independent.

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed (b) For any irreducible variety V /K, the valuation topologies on V (K) are independent. (c) The valuation topologies on A1(K) = K 1 are independent. (c’) i = j = ⇒ OiOj = K

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed (b) For any irreducible variety V /K, the valuation topologies on V (K) are independent. (c) The valuation topologies on A1(K) = K 1 are independent. (c’) i = j = ⇒ OiOj = K (d) For any irreducible curve C/K the valuation topologies on C(K) are independent.

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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E.c. multi-valued fields

In more detail,

Lemma

The following are equivalent for a multi-valued field (K, O1, . . . , On) with K = K alg and Oi = K: (a) K is existentially closed (b) For any irreducible variety V /K, the valuation topologies on V (K) are independent. (c) The valuation topologies on A1(K) = K 1 are independent. (c’) i = j = ⇒ OiOj = K (d) For any irreducible curve C/K the valuation topologies on C(K) are independent. One shows (c) = ⇒ (d) = ⇒ (a) = ⇒ (b) = ⇒ (c) ⇐ ⇒ (c’).

Will Johnson Multi-valued fiels March 4, 2018 5 / 23

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Failure of QE and NIP

Consider the theory of algebraically closed fields of characteristic = 2, with two independent valuations O1, O2. Let mi denote the maximal ideal of Oi. For i = 1, 2, let si : 1 + mi → 1 + mi be the inverse of the squaring map. If x ∈ 1 + m1 ∩ m2, then s1(x) = ±s2(x).

Will Johnson Multi-valued fiels March 4, 2018 6 / 23

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Failure of QE and NIP

Consider the theory of algebraically closed fields of characteristic = 2, with two independent valuations O1, O2. Let mi denote the maximal ideal of Oi. For i = 1, 2, let si : 1 + mi → 1 + mi be the inverse of the squaring map. If x ∈ 1 + m1 ∩ m2, then s1(x) = ±s2(x). Consider Q(i) with the (1 − 2i)-adic and (1 + 2i)-adic valuations. Then s1(−4) = 2i = −2i = s2(−4) Consider Q(i) with the (1 − 2i)-adic and (1 − 2i)-adic valuations. Then s1(−4) = 2i = 2i = s2(−4) The substructure generated by −4 is the same in the preceding two examples, so quantifier elimination fails.

Will Johnson Multi-valued fiels March 4, 2018 6 / 23

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Failure of QE and NIP

If ǫ1, ǫ2, . . . is a pairwise-distinct sequence in m1 ∩ m2, it turns out

  • ne can always find an x such that

s1(x + ǫi) = (−1)is2(x + ǫi) Taking the ǫi to be indiscernible, NIP fails. A similar argument works in characteristic 2. Algebraically closed fields with two valuations are never NIP.

Will Johnson Multi-valued fiels March 4, 2018 7 / 23

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Digression: an interesting consequence

Observation (various people)

The following statements are equivalent: (a) Every strongly dependent valued field is henselian. (b) No strongly dependent field defines two independent valuations. (c) No strongly dependent field defines two incomparable valuations. Conjecturally, all these statements are true. The implication (a) = ⇒ (b) uses the previous slide.

Will Johnson Multi-valued fiels March 4, 2018 8 / 23

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From independent valuations to arbitrary valuations

The theory of algebraically closed fields with n independent valuations has good model theory Model-completeness A weak form of quantifier elimination

Will Johnson Multi-valued fiels March 4, 2018 9 / 23

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From independent valuations to arbitrary valuations

The theory of algebraically closed fields with n independent valuations has good model theory Model-completeness A weak form of quantifier elimination How do we generalize to arbitrary valuations?

Will Johnson Multi-valued fiels March 4, 2018 9 / 23

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The tree of valuation rings on a field

Fix a field K. Let P be the poset of valuation rings on K. Then P has the following properties: P is a ∨-semilattice, with O1 ∨ O2 = O1 · O2 P has a maximal element K. For any a ∈ P, the set {x ∈ P|x ≥ a} is totally ordered. We will call such a poset a tree poset.

Will Johnson Multi-valued fiels March 4, 2018 10 / 23

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The tree of valuation rings on a field

Fix a field K. Let P be the poset of valuation rings on K. Then P has the following properties: P is a ∨-semilattice, with O1 ∨ O2 = O1 · O2 P has a maximal element K. For any a ∈ P, the set {x ∈ P|x ≥ a} is totally ordered. We will call such a poset a tree poset.

Remark

If S is a finite subset of P, the upper-bounded ∨-semilattice generated by S is a finite tree poset.

Will Johnson Multi-valued fiels March 4, 2018 10 / 23

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Prescribing a hierarchy of valuation rings

Theorem

Fix a finite tree poset (P, ∨, 1). Consider structures (K, Oa : a ∈ P) consisting of a field K and a valuation ring Oa for each a ∈ P. Consider the following theories: T 0

P asserts that O1 = K and the map a → Oa is weakly

  • rder-preserving.

TP asserts that K = K alg and the map a → Oa is a strictly

  • rder-preserving homomorphism of upper-bounded ∨-semilattices.

Then TP is the model companion of T 0

P.

Will Johnson Multi-valued fiels March 4, 2018 11 / 23

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Prescribing a hierarchy of valuation rings

Theorem

Fix a finite tree poset (P, ∨, 1). Consider structures (K, Oa : a ∈ P) consisting of a field K and a valuation ring Oa for each a ∈ P. Consider the following theories: T 0

P asserts that O1 = K and the map a → Oa is weakly

  • rder-preserving.

TP asserts that K = K alg and the map a → Oa is a strictly

  • rder-preserving homomorphism of upper-bounded ∨-semilattices.

Then TP is the model companion of T 0

P.

Remark

Up to definable expansions, every multi-valued algebraically closed field is a model of TP for appropriately chosen P.

Will Johnson Multi-valued fiels March 4, 2018 11 / 23

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Prescribing a hierarchy of valuation rings

Fix a finite tree poset (P, ∨, 1). Let a1, . . . , an enumerate the maximal elements of P \ {1}. Let Pi = {x ∈ P|x ≤ ai}. Note that each Pi is a finite tree poset.

Will Johnson Multi-valued fiels March 4, 2018 12 / 23

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Prescribing a hierarchy of valuation rings

Fix a finite tree poset (P, ∨, 1). Let a1, . . . , an enumerate the maximal elements of P \ {1}. Let Pi = {x ∈ P|x ≤ ai}. Note that each Pi is a finite tree poset. A model of T 0

P can be thought of as a field K with valuation rings

O1, . . . , On, and a T 0

Pi structure on the ith residue field ki.

Will Johnson Multi-valued fiels March 4, 2018 12 / 23

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Prescribing a hierarchy of valuation rings

Fix a finite tree poset (P, ∨, 1). Let a1, . . . , an enumerate the maximal elements of P \ {1}. Let Pi = {x ∈ P|x ≤ ai}. Note that each Pi is a finite tree poset. A model of T 0

P can be thought of as a field K with valuation rings

O1, . . . , On, and a T 0

Pi structure on the ith residue field ki.

Such a structure is a model of TP if (K, O1, . . . , On) is existentially closed and each residue field is a model of TPi.

Will Johnson Multi-valued fiels March 4, 2018 12 / 23

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Multi-valued fields with residue structure

For i = 1, . . . , n let Ti be a model-complete 1-sorted expansion of ACF. Let T be the theory of (n + 1)-sorted structures (K, k1, . . . , kn), with A field structure on K A residue map K ki for each i A (Ti)∀-structure on each ki.

Will Johnson Multi-valued fiels March 4, 2018 13 / 23

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Multi-valued fields with residue structure

For i = 1, . . . , n let Ti be a model-complete 1-sorted expansion of ACF. Let T be the theory of (n + 1)-sorted structures (K, k1, . . . , kn), with A field structure on K A residue map K ki for each i A (Ti)∀-structure on each ki.

Lemma

A model (K, k1, . . . , kn) | = T is e.c. exactly when the following conditions hold: K = K alg Each Oi is non-trivial and the Oi are pairwise-independent. ki | = Ti for all i.

Will Johnson Multi-valued fiels March 4, 2018 13 / 23

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Amalgamation over algebraically closed bases

Fix a finite tree poset P.

Theorem

In the category of models of T 0

P, the amalgamation problem

K0

  • K1

K2 can be solved whenever K0 = K alg .

Will Johnson Multi-valued fiels March 4, 2018 14 / 23

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Proof of amalgamation

By induction using the following:

Lemma

Let K0

  • K1
  • K2

K3

be a diagram of fields such that K0 = K alg and K1 ⊗K0 K2 injects into K3. Let O1, O2 be valuation rings on K1, K2 having the same restriction to K0. Then there is O3 on K3 extending O1 and O2. Moreover, O3 can be chosen so that res(K1) ⊗res(K0) res(K2) ֒ → res(K3) is injective.

Will Johnson Multi-valued fiels March 4, 2018 15 / 23

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Corollaries of amalgamation

Corollary

If K = K alg | = T 0

P, then K has the same type when embedded into any

model of TP.

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Corollaries of amalgamation

Corollary

If K = K alg | = T 0

P, then K has the same type when embedded into any

model of TP.

Corollary

TP has elimination of quantifiers “up to algebraic covers.”

Will Johnson Multi-valued fiels March 4, 2018 16 / 23

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Corollaries of amalgamation

Corollary

If K = K alg | = T 0

P, then K has the same type when embedded into any

model of TP.

Corollary

TP has elimination of quantifiers “up to algebraic covers.”

Corollary

The theory TP is decidable. More generally, the theory of n-multivalued algebraically closed fields is decidable.

Will Johnson Multi-valued fiels March 4, 2018 16 / 23

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Probable truth

Fix K | = T 0

P, and let ϕ(

a) be a TP-formula with parameters a ∈ K. By almost-q.e., there is a finite normal extension L/K such that, in models of TP extending K, the truth of ϕ( a) is determined by how the valuations are extended to L.

Will Johnson Multi-valued fiels March 4, 2018 17 / 23

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Probable truth

Fix K | = T 0

P, and let ϕ(

a) be a TP-formula with parameters a ∈ K. By almost-q.e., there is a finite normal extension L/K such that, in models of TP extending K, the truth of ϕ( a) is determined by how the valuations are extended to L. There are finitely many ways to extend the T 0

P-structure from K to L.

Consider the uniform distribution on this set.

Will Johnson Multi-valued fiels March 4, 2018 17 / 23

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Probable truth

Fix K | = T 0

P, and let ϕ(

a) be a TP-formula with parameters a ∈ K. By almost-q.e., there is a finite normal extension L/K such that, in models of TP extending K, the truth of ϕ( a) is determined by how the valuations are extended to L. There are finitely many ways to extend the T 0

P-structure from K to L.

Consider the uniform distribution on this set. Let P(ϕ( a)|K) denote the probability that ϕ( a) holds in a model of TP extending a random extension of the T 0

P-valuations to L.

Will Johnson Multi-valued fiels March 4, 2018 17 / 23

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Probable truth

Fix K | = T 0

P, and let ϕ(

a) be a TP-formula with parameters a ∈ K. By almost-q.e., there is a finite normal extension L/K such that, in models of TP extending K, the truth of ϕ( a) is determined by how the valuations are extended to L. There are finitely many ways to extend the T 0

P-structure from K to L.

Consider the uniform distribution on this set. Let P(ϕ( a)|K) denote the probability that ϕ( a) holds in a model of TP extending a random extension of the T 0

P-valuations to L.

This is independent of the choice of L.

Will Johnson Multi-valued fiels March 4, 2018 17 / 23

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Probable truth: key properties

P(−|K) defines a measure on the type-space of embeddings of K into models of TP.

Will Johnson Multi-valued fiels March 4, 2018 18 / 23

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Probable truth: key properties

P(−|K) defines a measure on the type-space of embeddings of K into models of TP. Probable truth is automorphism invariant.

Will Johnson Multi-valued fiels March 4, 2018 18 / 23

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Probable truth: key properties

P(−|K) defines a measure on the type-space of embeddings of K into models of TP. Probable truth is automorphism invariant. Let L/K be an extension of models of T 0

  • P. Suppose that for every

a ∈ P, the extension of residue fields with respect to Oa is relatively algebraically closed. Then P(ϕ( b)|L) = P(ϕ( b)|K) for every formula ϕ and tuple b ∈ K.

Will Johnson Multi-valued fiels March 4, 2018 18 / 23

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NTP2

Fix a finite tree poset P and let N be the number of leaves in the tree.

Theorem

In the theory TP, the home sort has burden at most 2N. In other words, there does not exist a model M | = TP, a formula ϕ(x; y), and an array ϕ(x; b1,1), ϕ(x; b1,2), ϕ(x; b1,3), · · · . . . ϕ(x; b2N+1,1), ϕ(x; b2N+1,2), ϕ(x; b2N+1,3), · · · with 2N + 1 rows and ω columns such that every row is k-inconsistent and every path η : [2N + 1] → ω through the rows is consistent.

Will Johnson Multi-valued fiels March 4, 2018 19 / 23

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NTP2

Fix a finite tree poset P and let N be the number of leaves in the tree.

Theorem

In the theory TP, the home sort has burden at most 2N. In other words, there does not exist a model M | = TP, a formula ϕ(x; y), and an array ϕ(x; b1,1), ϕ(x; b1,2), ϕ(x; b1,3), · · · . . . ϕ(x; b2N+1,1), ϕ(x; b2N+1,2), ϕ(x; b2N+1,3), · · · with 2N + 1 rows and ω columns such that every row is k-inconsistent and every path η : [2N + 1] → ω through the rows is consistent. By a result of Chernikov, it follows that TP is strong (hence NTP2).

Will Johnson Multi-valued fiels March 4, 2018 19 / 23

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Proof sketch

Part 1: find a locally indiscernible row.

Will Johnson Multi-valued fiels March 4, 2018 20 / 23

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Proof sketch

Part 1: find a locally indiscernible row. Extract a mutually indiscernible inp pattern.

Will Johnson Multi-valued fiels March 4, 2018 20 / 23

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Proof sketch

Part 1: find a locally indiscernible row. Extract a mutually indiscernible inp pattern. Choose an element a satisfying the 0th column.

Will Johnson Multi-valued fiels March 4, 2018 20 / 23

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Proof sketch

Part 1: find a locally indiscernible row. Extract a mutually indiscernible inp pattern. Choose an element a satisfying the 0th column. Consider each reduct (M, Op) for p ∈ P. As this reduct is dp-minimal, we can delete a row while making the remaining rows be mutually a-indiscernible in the reduct.

(See the proof that dp-rank is additive TODO)

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Proof sketch

Part 1: find a locally indiscernible row. Extract a mutually indiscernible inp pattern. Choose an element a satisfying the 0th column. Consider each reduct (M, Op) for p ∈ P. As this reduct is dp-minimal, we can delete a row while making the remaining rows be mutually a-indiscernible in the reduct.

(See the proof that dp-rank is additive TODO)

After running through all p ∈ P, at least one row ϕ(x; b0), ϕ(x; b1), . . . .

  • remains. This row is a-indiscernible in every reduct (M, Op).

Will Johnson Multi-valued fiels March 4, 2018 20 / 23

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Proof sketch

So far: a k-inconsistent sequence of formulas ϕ(x; b0), ϕ(x; b1), . . . such that b is a-indiscernible in every (M, Op). Also, M | = ϕ(a; b0).

Will Johnson Multi-valued fiels March 4, 2018 21 / 23

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Proof sketch

So far: a k-inconsistent sequence of formulas ϕ(x; b0), ϕ(x; b1), . . . such that b is a-indiscernible in every (M, Op). Also, M | = ϕ(a; b0). Find an algebraically closed base field B containing b such that P(ϕ(a; bi)|aB) = µ not dependent on i

Will Johnson Multi-valued fiels March 4, 2018 21 / 23

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Proof sketch

So far: a k-inconsistent sequence of formulas ϕ(x; b0), ϕ(x; b1), . . . such that b is a-indiscernible in every (M, Op). Also, M | = ϕ(a; b0). Find an algebraically closed base field B containing b such that P(ϕ(a; bi)|aB) = µ not dependent on i µ > 0 because ϕ(a; b0) is already true.

Will Johnson Multi-valued fiels March 4, 2018 21 / 23

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SLIDE 52

Proof sketch

So far: a k-inconsistent sequence of formulas ϕ(x; b0), ϕ(x; b1), . . . such that b is a-indiscernible in every (M, Op). Also, M | = ϕ(a; b0). Find an algebraically closed base field B containing b such that P(ϕ(a; bi)|aB) = µ not dependent on i µ > 0 because ϕ(a; b0) is already true. In a random extension of aB, ∼ k of the following formulas hold ϕ(a; b0), ϕ(a; b1), . . . , ϕ(a; b⌈k/µ⌉)

Will Johnson Multi-valued fiels March 4, 2018 21 / 23

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Proof sketch

So far: a k-inconsistent sequence of formulas ϕ(x; b0), ϕ(x; b1), . . . such that b is a-indiscernible in every (M, Op). Also, M | = ϕ(a; b0). Find an algebraically closed base field B containing b such that P(ϕ(a; bi)|aB) = µ not dependent on i µ > 0 because ϕ(a; b0) is already true. In a random extension of aB, ∼ k of the following formulas hold ϕ(a; b0), ϕ(a; b1), . . . , ϕ(a; b⌈k/µ⌉) By existential closure of M, and amalgamation over B, we can pull the situation back into M, contradicting k-inconsistency.

Will Johnson Multi-valued fiels March 4, 2018 21 / 23

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Open questions

If K is an algebraically closed field with n independent valuations, and if we add an NTP2 structure onto each residue field, is the resulting structure NTP2 as a whole?

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Open questions

If K is an algebraically closed field with n independent valuations, and if we add an NTP2 structure onto each residue field, is the resulting structure NTP2 as a whole? If K is a dp-minimal field and O1, . . . , On are arbitrary valuations on K, is (K, O1, . . . , On) strong?

Will Johnson Multi-valued fiels March 4, 2018 22 / 23

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Open questions

If K is an algebraically closed field with n independent valuations, and if we add an NTP2 structure onto each residue field, is the resulting structure NTP2 as a whole? If K is a dp-minimal field and O1, . . . , On are arbitrary valuations on K, is (K, O1, . . . , On) strong? If (K, . . .) is strong and O is arbitrary, must (K, . . . , O) be strong?

Will Johnson Multi-valued fiels March 4, 2018 22 / 23

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SLIDE 57

References

Model theory of fields with multiple valuations:

Lou van den Dries. “Model theory of Fields: Decidability and Bounds for Polynomial Ideals” 1978. Dissertation Yuri L. Ershov. Multi-valued Fields. Springer, 2001. Will Johnson. “Fun with fields” Chapter 11. 2016. Dissertation.

Background on dp-rank and burden

Artem Chernikov. “Theories without the Tree Property of the Second Kind.” Annals of pure and applied logic. Feb 2014. Itay Kaplan, Alf Onshuus, and Alexander Usvyatsov. “Additivity of the dp-rank.” Trans. Amer. Math. Soc. Nov 2013.

Will Johnson Multi-valued fiels March 4, 2018 23 / 23