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I have been reading about elementary embeddings in set theory and there is a question that has been nagging me:

Typically, one looks at elementary maps $j:V\to M$ with $M$ well-founded. Without any assumptions on the large cardinal strength of $V$, we cannot give examples of such maps. Also, typically, $j$ comes from an ultrapower construction.

If no non-principal ultrafilter is $\sigma$-complete, no ultrapower embedding $j:V\to M$ can even have $\omega^M$ standard (unless the ultrafilter is principal, of course).

Now, the question(s):

Is there any strength in the assumption that there is a (non-trivial) $j:V\to M$ elementary with $\omega^M$ standard?

It could be that this already ensures measurable cardinals, or it could be that the (consistency) strength increases with the well-foundedness of $M$ (I mean, with the standard part of $ORD^M$).

Note I am not assuming that $j$ comes from an ultrapower. If it matters, say the discussion takes places in NBG (or whatever is appropriate) so we can argue freely about classes.

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Does $\omega^M$ "standard" mean $\omega^M = \omega^V$? –  Oliver Feb 24 '11 at 3:12
    
If there is any nontrivial elementary embedding of $V$, then there exists a measurable cardinal. –  Oliver Feb 24 '11 at 3:16
1  
@Oliver: Yes for $\omega^M$ standard means that $\omega^M = \omega^V$. For your statement, this will not be true in general because you can get definable ultrapowers of the universe from ZFC for free by considering ultrapowers induced by general (finitely but not countably complete) ultrafilters. –  Jason Feb 24 '11 at 3:25
    
@Oliver: $\omega^M$ standard means that it is isomorphic to $\omega$. The result you quote only holds when the target is transitive: One can always get a nontrivial $j:V\to M$ by forming an ultrapower with respect to a non-principal ultrafilter (on $\omega$, if you wish), regardless of whether there are any large cardinals around. The $M$ we obtain in this case has non-standard $\omega$, but that's a different story. –  Bruce Feb 24 '11 at 3:29
    
@Jason: It looks like our comments overlapped! –  Bruce Feb 24 '11 at 3:29

1 Answer 1

up vote 6 down vote accepted

One natural interpretation of the question does give a measurable cardinal. Namely, suppose that $j:V\to M$ is an elementary embedding for which $M$ is an $\omega$-model. More precisely, we have a membership relation $E$ on $M$ and $j:\langle V,\in\rangle\to\langle M,E\rangle$ is an elementary embedding of these structures. Suppose that $j$ is nontrivial in the sense that it is not an isomorphism of every set with the $E$ predecessors of its image. In this case, I claim that there is a measurable cardinal. Fix any set $A$ for which $j$ is not an isomorphism of the elements of $A$ with the $E$-predecessors of $j(A)$. In this case, there must be an element $a\mathrel{E} j(A)$ which is not in the range of $j$. We may define a measure $\mu$ on $A$ by $X\in\mu\iff a\mathrel{E} j(X)$ for $X\subset A$. This is easily seen to be a nonprincipal ultrafilter on $A$. Furthermore, the fact that $M$ is an $\omega$-model will give us countable completeness of $\mu$, for if $X_n\in\mu$ for each $n$, the fact that $\omega^M$ is actually order type $\omega$ implies that every $E$ element of $\omega^M$ is $j(n)$ for some $n$, and from this it follows that the $E$ members of $j(\cap_n X_n)$ are precisely those in every $j(X_n)$. Thus, $a\in j(\cap_n X_n)$ and so $\cap_n X_n\in \mu$. So we have constructed a countably complete nonprincipal ultrafilter on a set $A$, and this implies the existence of a measurable cardinal. (The degree of completeness of $\mu$ is a measurable cardinal.)

Update. But there is another interpretation for which the property is weaker than measurability. Namely, it might happen that $j$ is an isomorphism of every set with the $E$-predecessors of its image, but $j$ is not an isomorphism of $\langle V,\in\rangle$ with $\langle M,E\rangle$. Thus, $j$ maps $V$ isomorphically to an initial segment of $M$, but $M$ has new objects at ranks above the ordinals of $V$. So essentially, $V$ is an elementary initial segment of $M$, with $j$ being essentially the inclusion map. This kind of situation can occur in general, since it is possible that $V_\delta\prec V_\kappa$, for example, when $\kappa$ is inaccessible, there are many such $\delta$. But for your question, we would have the situation where $M$ and $E$ are classes in $V$ that $V$ sees to be ill-founded. This is a more subtle situation, but one can build such an example starting only with a weakly compact cardinal, which is strictly weaker than measurability in consistency strength. Suppose $\kappa$ is weakly compact. Let $M_0$ be a transitive set of size $\kappa$, with $\kappa\in M$ and $V_\kappa\subset M$. Since $\kappa$ is weakly compact, there is an elementary embedding $j_0:M_0\to M_1$ to a transitive set $M_1$ of size $\kappa$ with critical point $\kappa$. Applying weak compactness again, we get a map $j_1:M_1\to M_2$, again with critical point $\kappa$. Iterating this, we build $j_n:M_n\to M_{n+1}$ with critical point $\kappa$ each time. Thus, we get an elementary map $j$ from $M_0$ to the direct limit model, which is a structure $\langle N,E\rangle$ that includes $V_\kappa$ in its well-founded part, but which is ill-founded above $\kappa$ and indeed, it has no $\kappa$-th element (since the critical point was $\kappa$ each time, new elements were inserted below any given element of the thread above $\kappa$). The ill-foundedness of the model exists below $j(\kappa)$. Thus, the structure $\langle V_\kappa,\in\rangle$ has an elementary extension to the $j(\kappa)$-rank initial segment of the direct limit. That is, we have $\langle V_\kappa,\in\rangle\prec\langle M,E\rangle$. The structure $M$ and relation $E$ on $M$ have size $\kappa$ and can therefore be coded using subsets of $\kappa$. But now the key point is that since $\kappa$ is inaccessible, we may equip $\langle V_\kappa,\in\rangle$ with all its subsets and still have a GBC model, indeed, a model of Kelly-Morse set theory. This model can see the elementary extension of $V_\kappa$ to the ill-founded model $M$, which is well-founded up to the height of $V_\kappa$. So this is an example of an elementary embedding of the $V$ of $V_\kappa$ into a class model that is an $\omega$-model, but still ill-founded, and is nontrivial in the sense that it is not an isomorphism, but one cannot extract any measurable cardinal, because it was built merely from a weakly compact cardinal.

(See the edit history for my earlier answer, which included some related information, which seems less relevant to me now.)

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Thank you, Dr. Hamkins. –  Bruce Feb 24 '11 at 6:02
    
Besides meaning $\kappa \in j(A)$ in my answer and not $j(\kappa) \in A$, my concern is the situation where we have all these sets that $M$ thinks are ordinals but really aren't. For example, what if $j(\kappa)$ contains a set $\{x\}$ for some $x \notin M$ that's not even an ordinal? Then $M$ thinks that $\{x\}$ is the empty set and so it thinks $\{x\} \in j(\kappa)$. Also, is it possible that the least $\alpha$ such that $j(\alpha) \neq \alpha$ is not even in $M$? –  Jason Feb 24 '11 at 9:19
    
Yes, Jason, my previous answer assumed that $M$ had an $\alpha$-th element, which was not warranted. I have fixed the argument to get around this. Any object in $j(\alpha)$ that is not in the range of $j$ will serve as a seed for a measure. I have edited. –  Joel David Hamkins Feb 24 '11 at 11:47
    
I streamlined the argument, since you don't even need to think about ordinals. –  Joel David Hamkins Feb 24 '11 at 12:36

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