I have some questions about the relationship from elementary substructures to elementary embeddings. It is clear how to go from elementary embeddings to substructures.

Given an nontrivial elementary embedding $\pi:V\to M$ ($\pi$ surjective, $M$ transitive) we have that $M$ is is an elementary substructure and we can define the critical point $\kappa:=\min\left\{ \alpha<On\mid\pi\left(\alpha\right)\neq\alpha\right\}$ and the ultrafilter $U:=\left\{ a\subseteq\kappa\mid\kappa\in\pi\left(a\right)\right\}$ .

Conversely, given a proper elementary substructure $M\subset V$ can I directly define $\kappa$ and $U$?

Another way of putting the question is if $\pi,\pi^{\prime}:V\to M$ are both elementary then must they both have the same critical point and ultrafilter? I would guess $\kappa=\left\{ \alpha<On\mid\alpha\notin M\right\}$ and so does not depend on $\pi$, but $U:=\left\{ a\subseteq\kappa\mid\kappa\in\pi\left(a\right)\right\}$ and $U^{\prime}:=\left\{ a\subseteq\kappa\mid\kappa\in\pi^{\prime}\left(a\right)\right\}$ do depend on $\pi$ and $\pi^{\prime}$ but are related somehow.

I believe I can use elementarity to prove the existence of elementary embeddings $V\to M$ and some of their properties. Suppose transitive $M$ is definable by statement $\varphi$ so the statement

$\zeta_{M} := \exists f\left(Function\left[f\right]\wedge\psi_{f,M}^{\prime}\wedge\psi_{f,M}^{\prime\prime}\wedge\psi_{f}^{\prime\prime\prime}\right)$


$\psi_{f,M}^{\prime} := \forall x\left(f\left(x\right)\in M\right)$

$\psi_{f,M}^{\prime\prime} := \forall x\left(x\in M\to f\left(x\right)=x\right)$

$\psi_{f}^{\prime\prime\prime} := \forall u,v\left(\left(u=v\leftrightarrow f\left(u\right)=f\left(v\right)\right)\wedge\left(u\in v\leftrightarrow f\left(u\right)\in f\left(v\right)\right)\right)$

can be expressed in 1st order set language.

Now $\zeta_{M}$ is witnessed by the identity function on $M$ so by elementarity $V\vDash\zeta_{M}$ so let $\pi$ witness $\zeta_{M}$ in V then $\pi:V\to M$ and is surjective by $\psi_{f,M}^{\prime}\wedge\psi_{f,M}^{\prime\prime}$ and preserves and reflects $=$ and $\in$ by $\psi_{f}^{\prime\prime\prime}$ (note reflecting $=$ makes $\pi$ injective). By $\psi_{f}^{\prime\prime\prime}$ and induction on the formation rules of statements I believe we can show $\pi$ is elementary.

If $\pi^{\prime}$ also witnesses $\zeta_{M}$ in $V$ then $\pi\left(x\right)=x=\pi^{\prime}\left(x\right)$ for $x\in M$ by $\psi_{f,M}^{\prime\prime}$. This gives $M\cap U=M\cap U^{\prime}$ because $a\in M\cap U$ iff $a\in M$ and $a\subseteq\kappa$ and $\kappa\in\pi\left(a\right)$ iff $a\in M$ and $a\subseteq\kappa$ and $\kappa\in\pi^{\prime}\left(a\right)$ iff $a\in M\cap U^{\prime}$.

If the above has logical flaws please help me understand them. If the above is correct, am I right about $\kappa$ being definable from $M$ only, but the ultrafilter depends on $\pi$, and if $\pi,\pi^{\prime}$ are two such embeddings their ultrafilters are related by $M\cap U=M\cap U^{\prime}$. What more can be said about all of this?


1 Answer 1


There several errors here. Most importantly are the two mistakes occurring when you claim

Given an nontrivial elementary embedding $\pi:V\to M$ ($\pi$ surjective, $M$ transitive) we have that $M$ is is an elementary substructure

First of all, your elementarity is going the wrong way. The image of $\pi$ - call it "$\pi[V]$" for simplicity - is an elementary substructure (well, subclassstructure) of $M$, but $M$ need not be an elementary substructure of $V$. Indeed, it usually won't be: e.g. taking $\pi$ to be the embedding coming from the first measurable, what $M$ thinks is the first measurable will in fact be much bigger than the first measurable.

EDIT: I missed the obvious fact that elementarity of this form can never happen: if $W$ is an inner model (= transitive proper class satisfying ZF), then $W\not\preccurlyeq V$ unless $W=V$. (Proof: by elementarity we have $(V_\alpha)^W=V_\alpha$ for every ordinal $\alpha$. Since $W$ is transitive we have $W=\bigcup_{\alpha\in Ord} (V_\alpha)^W=\bigcup_{\alpha\in Ord} V_\alpha=V$.)

Conversely, if $M\prec V$ there is no reason to believe that there is an elementary embedding from $V$ into $M$; rather, all we know is that the inclusion map is elementary from $M$ to $V$.

Secondly, $\pi$ is never surjective: the critical point of $\pi$ will never be in the image of $\pi$! If $\pi(a)=crit(\pi)$, by elementarity (and absoluteness) we must have $a$ be an ordinal; $\pi(crit(\pi))\not=\pi$, we have $a\not=crit(\pi)$, that is $\pi(a)\not=a$. But since $crit(\pi)$ is by definition the least ordinal moved by $\pi$, this tells us $a>crit(\pi)$. But now we have $$a>crit(\pi)\quad\mbox{ but }\quad \pi(a)<\pi(crit(\pi)),$$ contradicting elementarity.

A lesser issue: you also seem to assume that an elementary substructure is definable ("Suppose transitive $M$ is definable by statement $\varphi$"), which is in general not the case. (Of course, the image of an elementary embedding which is "set-generated" (e.g. coming from a measure) will be definable with a parameter; but note that that parameter will move when we apply the embedding!)

Finally, you claim

Now $\zeta_M$ is witnessed by the identity function on $M$

But this is not necessarily true! Just knowing that $M$ is definable in $V$ doesn't mean that the same definition pins down $M$ inside $M$. (That is, even if $M=\varphi^V$ we may still have $\varphi^M\not=M$.) The classic example of this is that while $L\models$ $"V=L,"$ we in general need not have $HOD\models$ $"V=HOD"$. And it should be clear that this is an issue, since if $M\subsetneq V$ then there is no class function from $V$ to $M$ which is the identity on $M$ and is injective so you can't build $\pi$ in $V$ as you suggest.

That said, there is an interesting question here once we clear away the issues mentioned above:

How much of an elementary embedding is determined by the target model $M$?

First, a very silly answer (working in ZF instead of ZFC for the moment): if there are two Reinhardt cardinals, then we know that the critical point can't be determined from the target model since both Reinhardt embeddings have target model $V$. Sorry, couldn't resist.

More seriously, even knowing the target model and the critical point won't pin down the embedding or even the corresponding measure (note that in general the embedding cannot be recovered from the induced measure!), even up to "reasonable equivalence;" for more details, look up the Mitchell order.

This leaves open the question of whether the critical point can be determined from the target. I believe the answer is no. As partial evidence for this, note that from any measurable $\kappa$ we can extract an elementary embedding of $L$ into itself with critical point $\kappa$. However, this doesn't quite count since the domain is $L$, rather than $V$ itself.

  • $\begingroup$ Thanks, so to clarify when we say $\pi:V\to M$ is an elementary embedding it is never the case that $\pi\left[V\right]=M$, so the role of $M$ is simply to provide a transitive class for the elementary substructure $\pi\left[V\right]$ to live in? $\endgroup$ Commented Mar 7, 2018 at 4:31
  • $\begingroup$ @MarkKortink Basically yes, but I worry that downplays $M$. Think about the usual measure situation. We have a measure $\mu$ on a cardinal $\kappa$. From $\mu$ we form as usual the ultrapower map $m: V\rightarrow\prod_\kappa V/\mu$, ignoring for the moment the coding issues in defining this precisely. Since $\mu$ is a measure, $\prod_\kappa V/\mu$ is well-founded (and it's clearly extensional and set-like) so we can apply the appropriate Mostowski collapse map $c$. The target model $M$ is the collapse of $\prod_\kappa V/\mu$, and our embedding $j$ is the composition of $c$ and $m$. (cont'd) $\endgroup$ Commented Mar 7, 2018 at 4:35
  • $\begingroup$ Now let's think about the elements of $M$ which aren't in the image of $j$. These are the $c$-images of the elements of $\prod_\kappa V/\mu$ which aren't in the image of $m$, and these in turn are exactly the $\mu$-classes which don't contain constant sequences! In other words, the "extra stuff" in $M$ is far from random: things in $M$ have "names" coming from $V$, and the map $j$ assigns to each thing in $V$ a very special name (in this setting, the corresponding constant sequence). So $M$ is more than just "something containing the image of $j$," it's really important. $\endgroup$ Commented Mar 7, 2018 at 4:37

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