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Alex Kruckman
Alex Kruckman
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  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with which fixes $M$ pointwise and has $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
added 29 characters in body
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Alex Kruckman
Alex Kruckman
  • 5k
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  • 24
  • 33
  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ withwhich fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.
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Alex Kruckman
Alex Kruckman
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First of all, the context of abstract elementary classes (AECs) is a generalization of the context of model theory of first-order logic. For any first-order theory $T$, we can let $K$ be the class of models of $T$, and let $\preceq$ be the elementary substructure relation between models of $T$. Then $(K,\preceq)$ is an AEC. [Thus the name: $(K,\preceq)$ is an elementary class, and the axioms of AECs are intended to abstract the properties of elementary classes].

For the rest of the answer, let's fix the first-order theory $T$, and the corresponding elementary class $(K,\preceq)$ as above.

Next, I'll explain how the notion of Galois type from AECs is a generalization of the notion of type (over a model) from model theory. Let $M\models T$. Let's write $S(M)$ for the set of first-order complete types over $M$ and $gS(M)$ for the set of Galois types over $M$ in $(K,\preceq)$. We'd like to establish a bijection between $S(M)$ and $gS(M)$. Let $p\in gS(M)$. Then $p$ is the $E_{K,\preceq}$ equivalence class of some triple $(M,N,a)$ where $M\preceq N$ and $a$ is a tuple from $N$. We map $p$ to $\text{tp}_N(a/M)\in S(M)$, the complete first-order type of $a$ over $M$ in the model $N$.

  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ with $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.

Ok, so when we're working with an elementary class, a Galois type over a model $M$ is essentially the same thing as a complete first-order type over $M$.

Finally, we want to see that the notion of saturation as defined by Jarden-Shelah is a generalization of the notion of saturation in model theory. In your quoted passage, Jarden-Shelah define what it means for $M$ to be "saturated in $\lambda^+$ over $\lambda$". I claim that when we're working with an elementary class, $M$ is saturated in $\lambda^+$ over $\lambda$ if and only if $M$ has cardinality $\lambda^+$ and is saturated (i.e. $\lambda^+$-saturated).

One direction is trivial: If $M$ has cardinality $\lambda^+$ and is $\lambda^+$-saturated, then for any model $N\preceq M$ with $|N| = \lambda$, every complete $n$-type over $N$ is realized in $M$ by saturation (i.e. $M$ is full over $N$), so $M$ is saturated in $\lambda^+$ over $\lambda$.

In the other direction, suppose $M$ is saturated in $\lambda^+$ over $\lambda$. Let $X\subseteq M$ with $|X| <\lambda^+$, and let $p(x)\in S(X)$ be a complete type over $X$. By Löwenheim-Skolem, we can pick an elementary substructure $N\preceq M$ with $X\subseteq N$ and $|N| = \lambda$. Let $p'(x)\in S(N)$ be any extension of $p(x)$ to a complete type over $N$. Since $M$ realizes every type in $S(N)$, $p'(x)$ is realized in $M$, and hence so is $p(x)$. So $M$ is $\lambda^+$-saturated.