15 Saturated Models

Definition 15.1. Let M be an infinite L-structure, and κ>|L|+0. We say M is κ-saturated if for any AM with |A|<κ, every type in SnM(A) is realised in M for all n.

Remark 15.2.

  • (i) Restricting to complete types is not important, as every n-type over A with respect to M can be extended to a complete type.
  • (ii) It suffices to assume n=1 to prove κ-saturation.
  • (iii) If M is κ-saturated then |M|κ. {xa:aM} is a consistent 1-type over M in M.

Definition 15.3 (Partial elementary / homogeneous). Let M,N be L-structures, AM, BN. A function f:AB is partial elementary if for every L-formulas φ(x1,,xn) and a1,,anA we have

Mφ(a¯)Nφ(f(a¯)).

Given κ|L|+0, M is κ-homogeneous if for any AM with |A|<κ, any partial elementary map fAM and any cM there is some dM with f{(c,a)} partial elementary. In other words, “partial elementary maps can be extended”.

For the rest of this section, assume T to be a complete L-theory with infinite models.

Definition 15.4. Define Sn(T)=SnM() for any / some MT (because if M,NT, then SnN()=SnM() as Th(M)=Th(N)=T).

Proposition 15.5. Assuming that:

  • T a complete L-theory with infinite models

  • MT

Then M is 0-saturated if and only if M is 0-homogeneous and M realises all types in Sn(T), n1.

Proof.

  • Assume M is 0-saturated. In particular, M realises all types in Sn(T) ( is finite).

    Fix some finite AM, and a partial elementary map f:AM. (Aim: find dM such that f{(c,d)} is partial elementary). Given cM, define pS1M(f(A)) to be such that

    φ(x,f(a¯))Mφ(c,a¯)

    (for all φ(y,x¯) L-formulas and a¯A). Write p=f(tpM(ca¯)).

    To show pS1M(f(A)), consider φ(x,a¯)p. Then Mφ(c,a¯), so Mx,φ(x,a¯), os Mφ(x,f(a¯)) as f is partial elementary.

    So p is finitely satisfiable, and completeness follows from tpM(cA) being complete. So as M is κ-saturated we have dp for some dM. Then f{(c,d)} is a partial elementary map.

  • Fix a1,,anM, pS1M({a1,,an}). We want to show p is realised in M.

    Set

    q={φ(y,x1,,xn):φ(y,a¯)p}Sn+1M(T).

    So by assumption, we have d,b1,,bn with (d,b¯)q.

    Consider tp(a¯)=tp(b¯). So f:biai is partial elementary. Let cM such that f{(d,c)} is partial elementary. Then tpM(c,a¯)=tpM(d,b¯). So (c,a¯)q and tp(ca¯)=p.

Notation. Given M, a¯,b¯Mn, write a¯Mb¯ if tpM(a¯)Mb¯.

So M is 0-homogeneous if and only if whenever a¯Mb¯ and cM, there exists dM with a¯cMb¯d.

Lemma 15.6. Assuming that:

  • T a complete L-theory with infinite models

  • MT

Then there is an NM with |N||M|+|L| and N is 0-homogeneous.

Proof. First claim: For any MT, there is NM with |N||M|+|L| and for any a¯,b¯,c from M such that a¯Mb¯ there is some dN with a¯cNb¯d.

Proof of claim: Enumerate all (a¯,b¯,c¯) as (a¯α,b¯α,c¯α)α|M|. Now let M0=M, and use transfinite induction to form a chain (Mα)α<|M|.

  • α is a limit ordinal: set Mα=i<αMi (then |Mα||α|(|M|+|L|)=|M|+|L| as α|M|).

  • Given α (not a limit ordinal), look at (aα,bα,cα). Assume a¯αMαb¯α so fα:a¯αb¯α partial elementary. Then we apply Proposition 13.4 (Note the elementary super structure construnted here is of size |M|+|L|) to find Mα+1Mα with |Mα+1||Mα|+|L||M|+|L| with a dMα+1 realising fα(tp(cαaα)). Then by construction, (a¯αc)Mα+1(b¯αd).

    Now let N=α<|M|Mα. Note that we might have introduced new elements. Note |N||M|(|M|+|L|)=|M|+|L|.

    We build a new chain M=N0N1N2 of countably many steps with |N1||M|+|L| and such that for any a¯,b¯,c¯N, if a¯Mb¯ then there is dNi+1 such that a¯cNi+1b¯d. We do this by iterating the claim.

    FInally let N=i<0Ni. Then:

    • |N||M|+|L|.

    • N is 0-homogeneous as any a¯,b¯,c from N lie in Ni for some i.

Definition 15.7 (Saturated). We say M is saturated if it is |M|-saturated.

Theorem 15.8. Assuming that:

  • T a complete L-theory with infinite models

  • L is countable

Then T has a countable saturated model if and only if Sn(T) is countable for every n1.

Proof.

  • MT countable, saturated.
    • Mn is countable for all n.

    • We have a map pa¯p (pSn(T), a¯ some realisation). This is map since M saturated, and injective (because complete types).

    So Sn(T) is countable.

  • Enumerate n1Sn(T)={p1,p2,p3,}. Fix a countable M0T, and build a chain M0M1M2 such that Mi realises pi and is countable, using Proposition 13.4.

    Get N=iMi. Then NT and is countable. N realises all types over .

    Apply Lemma 15.6 to get MN countable and 0-homogeneous structure.

    So by Proposition 15.5 is is 0-saturated.

Example 15.9.

  • (i) Let T=ACFp and let
    F={if p=0𝔽pif p>.

    Then

    Sn(T)=Spec(F[x1,,xn]).

    Thus Sn(T) is countable, since every ideal in Spec(F[x1,,xn]) is finitely generated (Hilbert’s basis theorem). So by Theorem 15.8, ACFp has a countable saturated model which is the model of transcendence degree 0.

    Note: if FACFp has transcendence degree n, the type determined by “x1,,xn+1” is an algebraically independent set.

  • (ii) Let T=TFDAG (torsion free, divisible abelian groups). This has a countable saturated model, which is the -vector space of dimension 0.
  • (iii) Let T=Th(,+,0). For n1, let δn(x) be the L-formula y(x=ny), and let be the set of primes. Given X finite,
    qx={δn(x):nX}{¬δn(x):nX}.

    qX is satisfiable in , thus pXS1(T) with qxpx.

    If XY, then pXpY, so |S1(T)|20. By Theorem 15.8, T doesn’t have a countable saturated model.

Example 15.10. Let MRG. We describe S1M(M). For aM, let paS1M(M) be the type containing “x=a” (exercise: why is this unique).

FOr VM set

pV={xa:aM}{F(x,q):aV}{¬E(x,a):aMV}.

pV is a 1-type with respect to M, not realised in M, determines a complete 1-type, as we have determined all atomic formula, by apV determines a complete type.

So S1M(M)={pa:aM}{pV:VM}.

|S1M(M)|=2|M|.

Note: in general |S1M(A)|2(|A|+|L|+0).

PIC

Proposition 15.11. Assuming that:

  • T a complete L-theory with infinite models

  • MN are both countable and saturated

Then MN.

Proof. Exercise (use back and forth argument).