[PPT] - Many-Sorted First-Order Model Theory Lecture 6 2 nd July, 2020 1 / PowerPoint Presentation

SLIDE 1

Many-Sorted First-Order Model Theory

Lecture 6 2nd July, 2020

1 / 37

SLIDE 2

Easy halves: unions of chains

Theorem 1 (Chang- Lo´ s-Suszko: easy direction)

Let (Ai)i<γ be a family of structures that form a chain under embedding. That is, the index set γ is an ordinal and i ≤ j implies Ai ≤ Aj. Put A =

i<γ Ai. Let ϕ(x) be a Π2 formula, and a0 a tuple from A0. If

Ai | = ϕ(a0) for every i < γ, then A | = ϕ(a0).

Proof.

◮ Write ϕ explicitly, as ∀y · ∃z · ψ(x, y, z). ◮ Take any tuple b from A. Since b is finite it belongs to some Ai. ◮ Since Ai | = ϕ(a0) by assumption, we have Ai | = ∃z · ψ(a0, b, z). ◮ This is an existential formula with parameters from Ai, and Ai ≤ A. ◮ By the easy half of the Lo´ s-Tarski Theorem (on substructures), we have A | = ∃z · ψ(a0, b, z). ◮ As b was arbitrary, we have A | = ∀y · ∃z · ψ(a0, y, z), as claimed.

2 / 37

SLIDE 3

Easy halves: unions of chains

Theorem 1 (Chang- Lo´ s-Suszko: easy direction)

Let (Ai)i<γ be a family of structures that form a chain under embedding. That is, the index set γ is an ordinal and i ≤ j implies Ai ≤ Aj. Put A =

i<γ Ai. Let ϕ(x) be a Π2 formula, and a0 a tuple from A0. If

Ai | = ϕ(a0) for every i < γ, then A | = ϕ(a0).

Proof.

◮ Write ϕ explicitly, as ∀y · ∃z · ψ(x, y, z). ◮ Take any tuple b from A. Since b is finite it belongs to some Ai. ◮ Since Ai | = ϕ(a0) by assumption, we have Ai | = ∃z · ψ(a0, b, z). ◮ This is an existential formula with parameters from Ai, and Ai ≤ A. ◮ By the easy half of the Lo´ s-Tarski Theorem (on substructures), we have A | = ∃z · ψ(a0, b, z). ◮ As b was arbitrary, we have A | = ∀y · ∃z · ψ(a0, y, z), as claimed.

3 / 37

SLIDE 4

Unions of chains: algebraically closed fields

Example 2

Fix a prime p, and consider the chain GF(p) ≤ GF(p2) ≤ · · · ≤ GF(pi) ≤ · · · and let F be its union. For any n consider the sentence ϕn = ∀y · ∃x · ynxn + . . . y1x + y0 = 0. Note that ϕn is a Π2 sentence. ◮ We have GF(pi) | = ϕn for i ≥ n. ◮ Moreover, F =

j≥i GF(pj) for any i ∈ N.

◮ It follows that F | = ϕn for every n. ◮ Thus, F is an algebraically closed field of characteristic p.

Exercise 1

Let F be as above. Fill the gaps in the proof that F is an algebraically closed field of characteristic p.

4 / 37

SLIDE 5

Unions of chains: algebraically closed fields

Example 2

Fix a prime p, and consider the chain GF(p) ≤ GF(p2) ≤ · · · ≤ GF(pi) ≤ · · · and let F be its union. For any n consider the sentence ϕn = ∀y · ∃x · ynxn + . . . y1x + y0 = 0. Note that ϕn is a Π2 sentence. ◮ We have GF(pi) | = ϕn for i ≥ n. ◮ Moreover, F =

j≥i GF(pj) for any i ∈ N.

◮ It follows that F | = ϕn for every n. ◮ Thus, F is an algebraically closed field of characteristic p.

Exercise 1

Let F be as above. Fill the gaps in the proof that F is an algebraically closed field of characteristic p.

5 / 37

SLIDE 6

Exercises

Exercise 2

Let ϕ be the sentence ∃x, y · ∀z · ¬(x < z) ∨ ¬(z < y) in the language of a binary relation <. Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models (Cn)n∈N such that Cn | = ψ but

n∈N Cn |

= ψ. Conclude that Σ2 sentences are not preserved under unions of chains.

Exercise 3

Prove that positive formulas are preserved under onto homomorphisms.

Exercise 4 (Somewhat hard, but instructive)

Let {Ai : i < γ} be a chain of similar structures, and let A =

i<γ Ai.

Prove that A ∈ HSP{Ai : i < γ}.

Hint. Consider “eventually constant” sequences (ui : i < γ), i.e., such if i > i0 (for some i0),

then ui = ai0, where ai0 ∈ |Ai0|. Next, consider the relation ∼ on these sequences, defined by (ui : i < γ) ∼ (wi : i < γ) if for all i > j0 (for some j0) we have ui = wi.

6 / 37

SLIDE 7

Exercises

Exercise 2

Let ϕ be the sentence ∃x, y · ∀z · ¬(x < z) ∨ ¬(z < y) in the language of a binary relation <. Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models (Cn)n∈N such that Cn | = ψ but

n∈N Cn |

= ψ. Conclude that Σ2 sentences are not preserved under unions of chains.

Exercise 3

Prove that positive formulas are preserved under onto homomorphisms.

Exercise 4 (Somewhat hard, but instructive)

Let {Ai : i < γ} be a chain of similar structures, and let A =

i<γ Ai.

Prove that A ∈ HSP{Ai : i < γ}.

Hint. Consider “eventually constant” sequences (ui : i < γ), i.e., such if i > i0 (for some i0),

then ui = ai0, where ai0 ∈ |Ai0|. Next, consider the relation ∼ on these sequences, defined by (ui : i < γ) ∼ (wi : i < γ) if for all i > j0 (for some j0) we have ui = wi.

7 / 37

SLIDE 8

Exercises

Exercise 2

Let ϕ be the sentence ∃x, y · ∀z · ¬(x < z) ∨ ¬(z < y) in the language of a binary relation <. Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models (Cn)n∈N such that Cn | = ψ but

n∈N Cn |

= ψ. Conclude that Σ2 sentences are not preserved under unions of chains.

Exercise 3

Prove that positive formulas are preserved under onto homomorphisms.

Exercise 4 (Somewhat hard, but instructive)

Let {Ai : i < γ} be a chain of similar structures, and let A =

i<γ Ai.

Prove that A ∈ HSP{Ai : i < γ}.

Hint. Consider “eventually constant” sequences (ui : i < γ), i.e., such if i > i0 (for some i0),

then ui = ai0, where ai0 ∈ |Ai0|. Next, consider the relation ∼ on these sequences, defined by (ui : i < γ) ∼ (wi : i < γ) if for all i > j0 (for some j0) we have ui = wi.

8 / 37

SLIDE 9

Preservation under substructures

Theorem 3 ( Lo´ s-Tarski)

Let T be a theory. If T is preserved under substructures, then T is equivalent to a set of Π1 formulas.

Proof.

◮ Wlog, T is consistent. Let T∀ be {ϕ ∈ Π1 : T | = ϕ}. ◮ Let K = {A: A ≤ B for some B ∈ Mod(T)}. So K is the class of submodels of models of T. ◮ As T is preserved by substructures, we have Mod(T) = K ⊆ Mod(T∀). We will show that Mod(T∀) ⊆ K. ◮ Take A | = T∀. Claim A: diag(A) ∪ T is consistent. ◮ Take a finite D0 ⊆ diag(A) and a finite T0 ⊆ T. Put δ(a) = D0, where a are all diagram constants occurring in D0. ◮ If D0 ∪ T0 is inconsistent, then T0 | = ¬δ(a). Note that T0 does not mention a at all. So T0 entails ¬δ(a) for arbitrary a. ◮ So, T0 | = ∀x · ¬δ(x), and as δ is quantifier free, we have ∀x · ¬δ(x) ∈ T∀. ◮ Thus, in particular, A | = ¬δ(a). ◮ But δ(a) ∈ diag(A), so A | = δ(a). Contradiction. ◮ This proves Claim A.

9 / 37

SLIDE 10

Preservation under substructures

Theorem 3 ( Lo´ s-Tarski)

Let T be a theory. If T is preserved under substructures, then T is equivalent to a set of Π1 formulas.

Proof.

◮ Wlog, T is consistent. Let T∀ be {ϕ ∈ Π1 : T | = ϕ}. ◮ Let K = {A: A ≤ B for some B ∈ Mod(T)}. So K is the class of submodels of models of T. ◮ As T is preserved by substructures, we have Mod(T) = K ⊆ Mod(T∀). We will show that Mod(T∀) ⊆ K. ◮ Take A | = T∀. Claim A: diag(A) ∪ T is consistent. ◮ Take a finite D0 ⊆ diag(A) and a finite T0 ⊆ T. Put δ(a) = D0, where a are all diagram constants occurring in D0. ◮ If D0 ∪ T0 is inconsistent, then T0 | = ¬δ(a). Note that T0 does not mention a at all. So T0 entails ¬δ(a) for arbitrary a. ◮ So, T0 | = ∀x · ¬δ(x), and as δ is quantifier free, we have ∀x · ¬δ(x) ∈ T∀. ◮ Thus, in particular, A | = ¬δ(a). ◮ But δ(a) ∈ diag(A), so A | = δ(a). Contradiction. ◮ This proves Claim A.

10 / 37

SLIDE 11

Preservation under substructures

Proof continued.

◮ By Claim A diag(A) ∪ T has a model, say, M. ◮ Since M | = diag(A), by the diagram lemma A ≤ M. ◮ So, A ∈ K as claimed, finishing the proof.

◮ The “arbitrary constant” trick is a formal version of a common practice of proving a general statement by picking some arbitrary elements.

Exercise 5 (Very easy, but instructive)

Formally, the arbitrary constant trick is the following statement. ◮ Let Σ be a signature, and C a set of new constants. Let S be a set of Σ[C]-sentences in which no constant from c occurs, or, which amounts to the same thing, a set of Σ-sentences in the signature Σ[C]. Let ϕ(c) be a Σ[C]-sentence, with c a sequence of constants from C. Then, S | = ϕ(c) implies S | = ∀x · ϕ(x). Prove it without recourse to completeness, soundness, or proof rules.

11 / 37

SLIDE 12

Preservation under substructures

Proof continued.

◮ By Claim A diag(A) ∪ T has a model, say, M. ◮ Since M | = diag(A), by the diagram lemma A ≤ M. ◮ So, A ∈ K as claimed, finishing the proof.

◮ The “arbitrary constant” trick is a formal version of a common practice of proving a general statement by picking some arbitrary elements.

Exercise 5 (Very easy, but instructive)

Formally, the arbitrary constant trick is the following statement. ◮ Let Σ be a signature, and C a set of new constants. Let S be a set of Σ[C]-sentences in which no constant from c occurs, or, which amounts to the same thing, a set of Σ-sentences in the signature Σ[C]. Let ϕ(c) be a Σ[C]-sentence, with c a sequence of constants from C. Then, S | = ϕ(c) implies S | = ∀x · ϕ(x). Prove it without recourse to completeness, soundness, or proof rules.

12 / 37

SLIDE 13

Preservation under substructures

Proof continued.

◮ By Claim A diag(A) ∪ T has a model, say, M. ◮ Since M | = diag(A), by the diagram lemma A ≤ M. ◮ So, A ∈ K as claimed, finishing the proof.

◮ The “arbitrary constant” trick is a formal version of a common practice of proving a general statement by picking some arbitrary elements.

Exercise 5 (Very easy, but instructive)

Formally, the arbitrary constant trick is the following statement. ◮ Let Σ be a signature, and C a set of new constants. Let S be a set of Σ[C]-sentences in which no constant from c occurs, or, which amounts to the same thing, a set of Σ-sentences in the signature Σ[C]. Let ϕ(c) be a Σ[C]-sentence, with c a sequence of constants from C. Then, S | = ϕ(c) implies S | = ∀x · ϕ(x). Prove it without recourse to completeness, soundness, or proof rules.

13 / 37

SLIDE 14

Arbitrary constant trick in our formal setting

S[c] S[x] Σ[c]

i

Σ[x]

Σ

ιc

ιx
T

The diagram on the left preserves Σ (i.e. i : Σ[c] → Σ[x] is the identity on Σ) and i maps c to x.

1. In our formal setting a variable is a triple, so we

have x = (x, s, Σ). If one ignores the third component (which is what we do here, to keep things as classical as we can), then both Sen(ιc) : Sen(Σ) → Sen(Σ[c]) and Sen(ιx) : Sen(Σ) → Sen(Σ[x]) are inclusions.

2. Since we work under the assumption that the

carrier sets of the models are non-empty sets, all injective signature morphisms are conservative; in particular, inclusions are conservative; By the two remarks above, it is safe to ignore the signature Σ, and write, simply, (a) T ∪ {S[x]} instead of ιc(T) ∪ {S[x]}, and (b) T ∪ {S[x]} ⊢ ⊥ instead of T ∪ {S[x]} ⊢Σ[x] ⊥. Now, if T ∪ {S[c]} is inconsistent, then since i(T ∪ {S[c]}) = T ∪ {S[x]}, by (Translation), T ∪ {S[x]} is inconsistent as well; since T ∪ S[x] ⊢ ⊥, by (NegI ), T ⊢ ¬S[x]; by the rule of generalization, we get T ⊢ ∀x · ¬S[x]; by soundness, T | = ∀x · ¬S[x].

14 / 37

SLIDE 15

Elementary substructures, elementary embeddings, and elementary equivalence

15 / 37

SLIDE 16

Elementary substructures and embeddings

Definition 4

Let A and B be Σ-structures. Let f : A → B be a homomorphism. ◮ A is an elementary substructure of B, (written A B) if A ≤ B and, for every formula ϕ(x) and every tuple a from |A|, we have A | = ϕ(a) iff B | = ϕ(a). ◮ f : A → B is an elementary embedding (written f : A

e

֒ → B) if f is an embedding (written f : A ֒ → B) and for every formula ϕ(x) and every tuple a from |A|, we have A | = ϕ(a) iff B | = ϕ(f (a)). ◮ If there exists an elementary embedding of A into B, we write A B.

Example 5

Let A and B be infinite pure identity structures. Then ◮ “A B and B A imply A ∼ = B” holds trivially (in fact, A = B). ◮ “A B and B A imply A ∼ = B” is Schr¨

der-Bernstein Theorem.

16 / 37

SLIDE 17

Elementary substructures and embeddings

Definition 4

Let A and B be Σ-structures. Let f : A → B be a homomorphism. ◮ A is an elementary substructure of B, (written A B) if A ≤ B and, for every formula ϕ(x) and every tuple a from |A|, we have A | = ϕ(a) iff B | = ϕ(a). ◮ f : A → B is an elementary embedding (written f : A

e

֒ → B) if f is an embedding (written f : A ֒ → B) and for every formula ϕ(x) and every tuple a from |A|, we have A | = ϕ(a) iff B | = ϕ(f (a)). ◮ If there exists an elementary embedding of A into B, we write A B.

Example 5

Let A and B be infinite pure identity structures. Then ◮ “A B and B A imply A ∼ = B” holds trivially (in fact, A = B). ◮ “A B and B A imply A ∼ = B” is Schr¨

der-Bernstein Theorem.

17 / 37

SLIDE 18

Tarski-Vaught test

Lemma 6 (Tarski-Vaught test)

Let A ≤ B be similar structures. If for every formula ϕ(x, a) with parameters a from A, we have that B | = ∃x · ϕ(x, a) implies A | = ∃x · ϕ(x, a), then A B holds.

Proof.

18 / 37

SLIDE 19

Tarski-Vaught test

Lemma 6 (Tarski-Vaught test)

Let A ≤ B be similar structures. If for every formula ϕ(x, a) with parameters a from A, we have that B | = ∃x · ϕ(x, a) implies A | = ∃x · ϕ(x, a), then A B holds.

Proof.

19 / 37

SLIDE 20

Unions of elementary chains

Lemma 7

Let {Ai : i < γ} be a family of similar structures such that Ai Aj for i ≤ j. Let C =

i<γ Ai. Then, Ai C holds for any i < γ.

Proof.

Exercise 6

◮ Prove that is an ordering relation on classes of similar structures. ◮ Give counterexamples showing that A ≤ B ≤ C and A C implies neither A B nor B C.

20 / 37

SLIDE 21

Unions of elementary chains

Lemma 7

Let {Ai : i < γ} be a family of similar structures such that Ai Aj for i ≤ j. Let C =

i<γ Ai. Then, Ai C holds for any i < γ.

Proof.

Exercise 6

◮ Prove that is an ordering relation on classes of similar structures. ◮ Give counterexamples showing that A ≤ B ≤ C and A C implies neither A B nor B C.

21 / 37

SLIDE 22

Unions of elementary chains

Lemma 7

Let {Ai : i < γ} be a family of similar structures such that Ai Aj for i ≤ j. Let C =

i<γ Ai. Then, Ai C holds for any i < γ.

Proof.

Exercise 6

◮ Prove that is an ordering relation on classes of similar structures. ◮ Give counterexamples showing that A ≤ B ≤ C and A C implies neither A B nor B C.

22 / 37

SLIDE 23

Elementary equivalence and isomorphism

Definition 8

Let A and B be Σ-structures. A and B are elementarily equivalent (written A ≡ B) if for every Σ-sentence ϕ we have A | = ϕ iff B | = ϕ.

Lemma 9

Let A and B be similar structures. Then, A ∼ = B implies A B and A B implies A ≡ B. The converses do not hold.

Exercise 7

Let A and B be infinite pure identity structures. Suppose |A| is a proper subset of |B|, of strictly smaller cardinality. ◮ Prove that A B (apply Tarski-Vaught test), but A ∼ = B. Conclude that B ≡ A, but B A. ◮ Show that A and B cannot be finite. Develop it into a proof of the fact that for finite structures elementary equivalence and isomorphism coincide.

23 / 37

SLIDE 24

Elementary equivalence and isomorphism

Definition 8

Let A and B be Σ-structures. A and B are elementarily equivalent (written A ≡ B) if for every Σ-sentence ϕ we have A | = ϕ iff B | = ϕ.

Lemma 9

Let A and B be similar structures. Then, A ∼ = B implies A B and A B implies A ≡ B. The converses do not hold.

Exercise 7

Let A and B be infinite pure identity structures. Suppose |A| is a proper subset of |B|, of strictly smaller cardinality. ◮ Prove that A B (apply Tarski-Vaught test), but A ∼ = B. Conclude that B ≡ A, but B A. ◮ Show that A and B cannot be finite. Develop it into a proof of the fact that for finite structures elementary equivalence and isomorphism coincide.

24 / 37

SLIDE 25

Elementary equivalence and isomorphism

Definition 8

Let A and B be Σ-structures. A and B are elementarily equivalent (written A ≡ B) if for every Σ-sentence ϕ we have A | = ϕ iff B | = ϕ.

Lemma 9

Let A and B be similar structures. Then, A ∼ = B implies A B and A B implies A ≡ B. The converses do not hold.

Exercise 7

Let A and B be infinite pure identity structures. Suppose |A| is a proper subset of |B|, of strictly smaller cardinality. ◮ Prove that A B (apply Tarski-Vaught test), but A ∼ = B. Conclude that B ≡ A, but B A. ◮ Show that A and B cannot be finite. Develop it into a proof of the fact that for finite structures elementary equivalence and isomorphism coincide.

25 / 37

SLIDE 26

Preservation under unions of chains

Theorem 10 (Chang- Lo´ s-Suszko)

Let T be a theory. If T is preserved under unions of chains, then T is equivalent to a set of Π2 formulas.

Proof.

◮ Let T∀∃ be {ϕ ∈ Π2 : T | = ϕ}. Thus, every model of T is a model of T∀∃. We will show that every model of T∀∃ is a model of T. ◮ Let A | = T∀∃. Expand the signature from Σ, to Σ[a], by naming all elements of A. ◮ Let D∀(A) = {ϕ ∈ Π1 : (A, a) | = ϕ}. D∀(A) is the set of all universal Σ[a]-sentences which are true in (A, a). ◮ Claim B: D∀(A) ∪ T is consistent. ◮ Suppose the contrary. Then for some D0 ⊆fin D∀(A) and T0 ⊆fin T, the set D0 ∪ T0 is inconsistent. Put δ(a0) = D0, where a0 are the constants occurring in D0. Written explicitly, δ(a0) is of the form ∀x · η(a0, x). ◮ Then, T0 | = ¬δ(a0), that is, T0 | = ¬∀x · η(a0, x). ◮ Therefore, T0 | = ∀y · ¬∀x · η(y, x), that is T0 | = ∀y · ∃x · ¬η(y, x). Arbitrary constant trick at work again. ◮ Thus, ∀y · ∃x · ¬η(y, x) ∈ T∀∃. Therefore, A | = ∀y · ∃x · ¬η(y, x). ◮ Thus, (A, a) | = ∃x · ¬η(a0, x). But ∃x · ¬η(a0, x) is ¬δ(a0), so (A, a) | = ¬δ(a0). ◮ On the other hand, δ(a0) = D0, so (A, a) | = δ(a0). Contradiction.

26 / 37

SLIDE 27

Preservation under unions of chains

Theorem 10 (Chang- Lo´ s-Suszko)

Let T be a theory. If T is preserved under unions of chains, then T is equivalent to a set of Π2 formulas.

Proof.

◮ Let T∀∃ be {ϕ ∈ Π2 : T | = ϕ}. Thus, every model of T is a model of T∀∃. We will show that every model of T∀∃ is a model of T. ◮ Let A | = T∀∃. Expand the signature from Σ, to Σ[a], by naming all elements of A. ◮ Let D∀(A) = {ϕ ∈ Π1 : (A, a) | = ϕ}. D∀(A) is the set of all universal Σ[a]-sentences which are true in (A, a). ◮ Claim B: D∀(A) ∪ T is consistent. ◮ Suppose the contrary. Then for some D0 ⊆fin D∀(A) and T0 ⊆fin T, the set D0 ∪ T0 is inconsistent. Put δ(a0) = D0, where a0 are the constants occurring in D0. Written explicitly, δ(a0) is of the form ∀x · η(a0, x). ◮ Then, T0 | = ¬δ(a0), that is, T0 | = ¬∀x · η(a0, x). ◮ Therefore, T0 | = ∀y · ¬∀x · η(y, x), that is T0 | = ∀y · ∃x · ¬η(y, x). Arbitrary constant trick at work again. ◮ Thus, ∀y · ∃x · ¬η(y, x) ∈ T∀∃. Therefore, A | = ∀y · ∃x · ¬η(y, x). ◮ Thus, (A, a) | = ∃x · ¬η(a0, x). But ∃x · ¬η(a0, x) is ¬δ(a0), so (A, a) | = ¬δ(a0). ◮ On the other hand, δ(a0) = D0, so (A, a) | = δ(a0). Contradiction.

27 / 37

SLIDE 28

Preservation under unions of chains

Proof continued: alternating chains method.

◮ Let B | = D∀(A) ∪ T. ◮ Since D∀(A) ⊇ diag(A), we have A ≤ B. ◮ Claim C: diag(B) ∪ Th(A, a) is consistent. ◮ Exercise. Prove it using the arbitrary constant trick. Note that any tuple b from |B| can be written as (c, d) with c ∈ |B| \ |A| and d ∈ |A|. ◮ Let A1 | = diag(B) ∪ Th(A, a). ◮ Then A ≤ B ≤ A1, and B | = T; moreover, A A1. The elementarity of the embedding follows from the fact that A1 | = Th(A, a). ◮ Continuing inductively, we get A ≤ B ≤ A1 ≤ B1 ≤ A2 ≤ . . . Draw a diagram! ◮ Let C be the union of this chain. ◮ Then, C is also the union of B ≤ B1 ≤ B2 ≤ . . . ◮ Since Bi | = T for every i by construction, and T is preserved under unions of chains, we get C | = T. ◮ But C is also the union of A ≤ A1 ≤ A2 ≤ . . . ◮ Since A A1 A2 . . . , by Lemma 7 we get A C. Exercise: there is a hidden use of a part of Exercise 6 here. Find it. ◮ Therefore, A | = T, as required.

28 / 37

SLIDE 29

Free algebras as basic/canonical models

Theorem 11 (Basic/Canonical model)

Let E be a set of atomic sentences in a signature Σ. Then, there exists a Σ-structure M such that

1. M |

= E.

2. Every element of |M| is of the form tM for a closed term t.
3. If N is an Σ-structure and N |

= E, then there exists a homomorphism h: M → N. If Σ is non-void, then this homomorphism is unique.

Proof.

By Theorem 26 from Lecture 2. We called M “basic”. It is often called “canonical”.

Example 12

Let G be the set of equations x · (y · z) = (x · y) · z, x−1 · x = e = x · x−1, x · e = x = e · x, that is, the group axioms. Let E be the set of all atomic sentences obtained from G by replacing variables by closed terms in the signature expanded by constants from some set X (here the “variables are constants” trick works very nicely). Then, the basic model G of X is the free group G[X] generated by X.

29 / 37

SLIDE 30

Free algebras as basic/canonical models

Theorem 11 (Basic/Canonical model)

Let E be a set of atomic sentences in a signature Σ. Then, there exists a Σ-structure M such that

1. M |

= E.

2. Every element of |M| is of the form tM for a closed term t.
3. If N is an Σ-structure and N |

= E, then there exists a homomorphism h: M → N. If Σ is non-void, then this homomorphism is unique.

Proof.

By Theorem 26 from Lecture 2. We called M “basic”. It is often called “canonical”.

Example 12

Let G be the set of equations x · (y · z) = (x · y) · z, x−1 · x = e = x · x−1, x · e = x = e · x, that is, the group axioms. Let E be the set of all atomic sentences obtained from G by replacing variables by closed terms in the signature expanded by constants from some set X (here the “variables are constants” trick works very nicely). Then, the basic model G of X is the free group G[X] generated by X.

30 / 37

SLIDE 31

Free algebras as basic/canonical models

Theorem 11 (Basic/Canonical model)

Let E be a set of atomic sentences in a signature Σ. Then, there exists a Σ-structure M such that

1. M |

= E.

2. Every element of |M| is of the form tM for a closed term t.
3. If N is an Σ-structure and N |

= E, then there exists a homomorphism h: M → N. If Σ is non-void, then this homomorphism is unique.

Proof.

By Theorem 26 from Lecture 2. We called M “basic”. It is often called “canonical”.

Example 12

Let G be the set of equations x · (y · z) = (x · y) · z, x−1 · x = e = x · x−1, x · e = x = e · x, that is, the group axioms. Let E be the set of all atomic sentences obtained from G by replacing variables by closed terms in the signature expanded by constants from some set X (here the “variables are constants” trick works very nicely). Then, the basic model G of X is the free group G[X] generated by X.

31 / 37

SLIDE 32

Adding roots of polynomials to a field

Example 13

Let K be a field. As usual we write K[x] for the ring of polynomials in x over K. We can think of K[x] as a structure in the signature of rings with additional constants for each element of K, and one more for x. Take a polynomial p(x) which is irreducible over K. Let T be the set of all equations that are true in K[x]. Now consider T ∪ {p(x) = 0}. This is a set of atomic sentences, so it has a basic model M. ◮ By diagram lemma, there is an onto homomorphism h: K[x] → M. In particular M is a ring. ◮ As M | = p(x) = 0, we have h(a) = 0M for every a in the ideal (p(x)]. ◮ Because K[x]p(x) satisfies T ∪ {p(x) = 0}, we get that K[x]p(x) is a homomorphic image of M. ◮ In fact, K[x]p(x) and M are isomorphic. For consider K[x] → M → K[x]p(x). Their composition is the quotient map.

32 / 37

SLIDE 33

Free algebras again

Let K be a class of similar algebras, and X a set of variables. We write EqX(K) for the set of all equations over X true in K. If X is clear from context we write Eq(K).

Lemma 14 (Free algebras)

Let K and X be as above, and E = EqX(K). Let F[X] be the basic model for E. Then, F[X] ∈ SP(K). Moreover, EqX(K) = EqX(F[X]).

Proof.

◮ Let Φ = {ϕi : i < γ} be the set of all homomorphisms ϕi : F[X] → Ki, for some Ki ∈ K. We take it up to identity of kernels, so it is a set. ◮ The map ψ : F[X] →

i<γ Ki, defined coordinatewise as ψ = (ϕi : i < γ), is a

homomorphism. ◮ By universal property of F[X], for every a, b ∈ |F[X]| with a = b, we have a homomorphism ϕi such that ϕi(a) = ϕi(b). Note that a and b are values of some terms t and s over X, so t ≈ s / ∈ E. ◮ Thus, the image ψ(F[X]) is a substructure of

i<γ Ki. Hence, F[X] ∈ SP(K), as claimed.

◮ For the moreover part: EqX (K) ⊆ EqX (F[X]) follows by preservation of equations by SP; EqX (K) ⊇ EqX (F[X]) follows by the remark above in blue.

33 / 37

SLIDE 34

Free algebras again

Let K be a class of similar algebras, and X a set of variables. We write EqX(K) for the set of all equations over X true in K. If X is clear from context we write Eq(K).

Lemma 14 (Free algebras)

Let K and X be as above, and E = EqX(K). Let F[X] be the basic model for E. Then, F[X] ∈ SP(K). Moreover, EqX(K) = EqX(F[X]).

Proof.

◮ Let Φ = {ϕi : i < γ} be the set of all homomorphisms ϕi : F[X] → Ki, for some Ki ∈ K. We take it up to identity of kernels, so it is a set. ◮ The map ψ : F[X] →

i<γ Ki, defined coordinatewise as ψ = (ϕi : i < γ), is a

homomorphism. ◮ By universal property of F[X], for every a, b ∈ |F[X]| with a = b, we have a homomorphism ϕi such that ϕi(a) = ϕi(b). Note that a and b are values of some terms t and s over X, so t ≈ s / ∈ E. ◮ Thus, the image ψ(F[X]) is a substructure of

i<γ Ki. Hence, F[X] ∈ SP(K), as claimed.

◮ For the moreover part: EqX (K) ⊆ EqX (F[X]) follows by preservation of equations by SP; EqX (K) ⊇ EqX (F[X]) follows by the remark above in blue.

34 / 37

SLIDE 35

Birkhoff’s HSP Theorem

Theorem 15 (Birkhoff)

Let K be a class of similar algebras, and X some countably infinite set of

variables. If HSP(K) ⊆ K, then K = Mod(EqX(K)).

Proof.

◮ K ⊆ Mod(EqX (K)) trivially holds. ◮ Let A ∈ Mod(EqX (K)). Fix some set Y , with card(Y ) = max{card(X), card(A)}. ◮ We have EqX (K) = EqY (K) (exercise). ◮ Let F[Y ] be the basic model (free algebra) for EqY (K). ◮ By Lemma 14, we have F[Y ] ∈ SP(K). ◮ By universal property of F[Y ] we have A ∈ H(F[Y ]) (since Y is large enough). ◮ Hence, A ∈ HSP(K), and so A ∈ K, as required.

Exercise 8

Prove that for any infinite X and Y we have EqX(K) = EqY (K).

35 / 37

SLIDE 36

Birkhoff’s HSP Theorem

Theorem 15 (Birkhoff)

Let K be a class of similar algebras, and X some countably infinite set of

variables. If HSP(K) ⊆ K, then K = Mod(EqX(K)).

Proof.

◮ K ⊆ Mod(EqX (K)) trivially holds. ◮ Let A ∈ Mod(EqX (K)). Fix some set Y , with card(Y ) = max{card(X), card(A)}. ◮ We have EqX (K) = EqY (K) (exercise). ◮ Let F[Y ] be the basic model (free algebra) for EqY (K). ◮ By Lemma 14, we have F[Y ] ∈ SP(K). ◮ By universal property of F[Y ] we have A ∈ H(F[Y ]) (since Y is large enough). ◮ Hence, A ∈ HSP(K), and so A ∈ K, as required.

Exercise 8

Prove that for any infinite X and Y we have EqX(K) = EqY (K).

36 / 37

SLIDE 37

Birkhoff’s HSP Theorem

Theorem 15 (Birkhoff)

Let K be a class of similar algebras, and X some countably infinite set of

variables. If HSP(K) ⊆ K, then K = Mod(EqX(K)).

Proof.

◮ K ⊆ Mod(EqX (K)) trivially holds. ◮ Let A ∈ Mod(EqX (K)). Fix some set Y , with card(Y ) = max{card(X), card(A)}. ◮ We have EqX (K) = EqY (K) (exercise). ◮ Let F[Y ] be the basic model (free algebra) for EqY (K). ◮ By Lemma 14, we have F[Y ] ∈ SP(K). ◮ By universal property of F[Y ] we have A ∈ H(F[Y ]) (since Y is large enough). ◮ Hence, A ∈ HSP(K), and so A ∈ K, as required.

Exercise 8

Prove that for any infinite X and Y we have EqX(K) = EqY (K).

37 / 37