15 Saturated Models
Definition 15.1.
Let
be an infinite -structure,
and .
We say
is -saturated
if for any
with ,
every type in
is realised in
for all .
Remark 15.2.
-
(i)
Restricting to complete types is not important, as every -type
over
with respect to
can be extended to a complete type.
-
(ii)
It suffices to assume
to prove -saturation.
-
(iii)
If
is -saturated
then .
is a consistent -type
over
in .
Definition 15.3 (Partial elementary / homogeneous).
Let
be
-structures,
,
. A function
is partial elementary
if for every -formulas
and
we have
Given ,
is -homogeneous
if for any
with ,
any partial elementary map
and any
there is some
with
partial elementary. In other words, “partial elementary maps can be extended”.
For the rest of this section, assume
to be a complete -theory
with infinite models.
Definition 15.4.
Define
for any / some
(because if ,
then
as ).
Proposition 15.5.
Assuming that:
Proof.
-
Assume
is -saturated.
In particular,
realises all types in
(
is finite).
Fix some finite , and a
partial elementary map .
(Aim: find such
that is partial
elementary). Given ,
define
to be such that
(for all
-formulas
and ).
Write .
To show ,
consider .
Then ,
so ,
os
as
is partial elementary.
So
is finitely satisfiable, and completeness follows from
being complete. So as
is -saturated
we have
for some .
Then
is a partial elementary map.
-
Fix ,
.
We want to show
is realised in .
Set
|
So by assumption, we have
with .
Consider .
So
is partial elementary. Let
such that
is partial elementary. Then .
So
and .
□
Notation.
Given ,
,
write
if .
So is
-homogeneous if and
only if whenever
and , there
exists
with .
Lemma 15.6.
Assuming that:
Then there is an
with
and
is
-
homogeneous.
Proof.
First claim: For any ,
there is
with
and for any
from
such that
there is some
with .
Proof of claim: Enumerate all
as . Now let
, and use transfinite
induction to form a chain .
-
is a limit ordinal: set
(then
as ).
-
Given
(not a limit ordinal), look at .
Assume
so
partial elementary. Then we apply Proposition 13.4 (Note the elementary super structure construnted
here is of size )
to find
with
with a
realising .
Then by construction, .
Now let .
Note that we might have introduced new elements. Note .
We build a new chain
of countably many steps with
and such that for any ,
if
then there is
such that .
We do this by iterating the claim.
FInally let .
Then:
□
Definition 15.7 (Saturated).
We say
is saturated if it is -saturated.
Theorem 15.8.
Assuming that:
Then has a countable
saturated model if and only if
is countable for every .
Proof.
-
countable, saturated.
-
is countable for all .
-
We have a map
(,
some realisation). This is map since
saturated, and injective (because complete types).
So
is countable.
-
Enumerate . Fix
a countable , and
build a chain
such that
realises
and is countable, using Proposition 13.4.
Get . Then
and is countable.
realises all
types over .
Apply Lemma 15.6 to get
countable and -homogeneous
structure.
So by Proposition 15.5 is is -saturated.
□
Example 15.9.
-
(i)
Let
and let
Then
|
Thus
is countable, since every ideal in
is finitely generated (Hilbert’s basis theorem). So by Theorem 15.8,
has a countable saturated model which is the model of transcendence degree .
Note: if
has transcendence degree ,
the type determined by “”
is an algebraically independent set.
-
(ii)
Let
(torsion free, divisible abelian groups). This has a countable saturated model, which is the
-vector
space of dimension .
-
(iii)
Let . For
, let
be the
-formula
, and let
be the set of
primes. Given
finite,
|
is satisfiable in ,
thus
with .
If ,
then ,
so .
By Theorem 15.8,
doesn’t have a countable saturated model.
Example 15.10.
Let .
We describe .
For ,
let
be the type containing “”
(exercise: why is this unique).
FOr set
|
is a -type
with respect to ,
not realised in ,
determines a complete -type,
as we have determined all atomic formula, by
determines a complete type.
So .
.
Note: in general .
Proposition 15.11.
Assuming that:
Then .
Proof.
Exercise (use back and forth argument). □