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Suppose $\kappa_0$ is a measurable cardinal and $\mu_0$ is a normal measure on $\kappa_0$. $M_1$ is the transitive collapse of $Ult(V,\mu_0)$, $j_{0,1}:V\rightarrow{M_1}$ is the elementary embedding induced by the ultrapower. In $M_1$, $\kappa_1=j_{0,1}(\kappa_0)$ is a measurable cardinal and $\mu_1$ is a normal measure on $\kappa_1$ in $M_1$ such that $\mu_1$ is not in the range of $j_{0,1}$. $M_2$ is the transitive collapse of $Ult(M_1,\mu_1)$, $j_{1,2}:M_1\rightarrow{M_2}$ is the elementary embedding induced by the ultrapower. $j_{0,2}=j_{1,2}\circ{j_{0,1}}$.

Is it true that: ``Suppose $N$ is an inner model, $i:V\rightarrow{N}$ and $k:N\rightarrow{M_2}$ are elementary embeddings such that $k\circ{i}=j_{0,2}$. Then $k''N=j_{0,2}''V$ or $k''N=j_{1,2}''M_1$ or $k''N=M_2$''?

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This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.

The answer is Yes.

Let me denote κ0 just by κ and j02 by j. Since μ1 is a measure in M1, it has the form j01(m)(κ), where m = (να | α < κ). Since you have said that μ1 is not in ran(j01), we may choose the να to be all different, and different from μ0. In this case, there is a partition of κ as the disjoint union of Xα, with Xα in να and none in μ0. Let x = (Xα | α < κ). Note that κ is not in j01(Xα) for any α < κ, and similarly κ1 is not in j(Xα). But κ is in j01(x)(β) for some β < κ1, since this is a partition of κ1. Apply j12 to conclude that κ1 is in j(x)(β) for this β. Thus, there is some β in the interval [κ, κ1) having the form β = j(f)(κ1) for the function f that picks the index. From this, it follows from normality of μ0 that we can write κ = j(g)(κ1) for some function g, since any β < κ1 generates κ via j01. In my favored terminology, the seed κ1 generates κ via j and in fact generates all β in [κ,κ1) via j.

Similarly, suppose that δ is in the interval [κ1,j(κ)). We know δ = j12(f)(κ1) for some function f on κ1 in M1. We also know f = j01(F)(κ) for some F in V. Thus, δ = j(F)(κ, κ1). In Y, let (α,β) be the smallest pair with δ = j(F)(α,β). It cannot be that both are below κ1, since this would be inside ran(j12) and so the least pair must have β = κ1. Thus, δ generates κ1, which we already observed generates κ.

To summarize, every ordinal in the interval [κ1,j(κ)) generates κ1, which generates all the ordinals β in [κ,κ1), any of which generate κ and all the other such β.

This is enough to answer your question. The k " N in your question is just an arbitrary elementary substructure of M2 containing ran(j), so suppose we have Y elementary in M2 and ran(j) subset Y. The case Y = ran(j) is one of your cases. Otherwise, Y has something not in ran(j). Every object in M2 has form j(h)(κ,κ1) for some function h, so by looking at the smallest pair of ordinals to generate a given object with j(h), we see that there must be ordinals below j(κ) in Y. If Y contains any ordinal δ in the interval [κ1,j(κ)), then it will contain both κ and κ1, since we observed that any such δ generates these ordinals. In this case, Y = M2, since those two ordinals generate everything. So we assume that Y contains no such δ. In this last case, Y must contain some ordinal β in the interval [κ,κ1). Since any such β generates κ, Y contains all such ordinals. It follows that ran(j12) subset Y and in fact = Y, since if Y contained anything more it would have to have an additional ordinal δ in [κ1,j(κ)).

So we've seen that your three cases are the only possibilities. And like your previous question, there is no need to assume that Y or N is somehow internally definable.

By the way, this was a problem that I had solved many years ago for my dissertation, although perhaps other people had also thought about it. I was interested in understanding which pairs of ordinals (α,β) generate product measures via an embedding j, and this question is very much related to that.

(Click the edit history to see my previous answer, which was just about the case when μ1 is in the range of j01, a case for which the answer is no.)

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I am the student exchange you with email, you introduce me to go here, so I register this new name. In my question, I require $\mu_1$ is not in the range of $j_{0,1}$, i.e. the second step of the iterated ultrapower is done by a normal measure not in $j_{0,1}''V$. For example, suppose $\kappa$ has $\kappa$ many normal measures <U_\alpha:\alpha<\kappa>, the first step of the iterated ultrapower I use $U_0$, and the second step of the iteration, I use the \kappa-th measure in the sequence j_{0,1}(<U_\alpha:\alpha<\kappa>) . –  Ant emyy Lee Feb 3 '10 at 3:20
    
So the model-theoretic fact you said in your answer may not make sense in this case, since $\mu_1$ is not a normal measure in $V$. How to deal with this case? –  Ant emyy Lee Feb 3 '10 at 3:20
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Welcome to MO! Ah, I read your question more carefully now. You are right: my answer applies to the case where mu_1 IS in the range of j_01, for then mu_1=j_01(nu), and in this case, j02 is precisely the ultrapower by mu_0 x nu, since every object in M_2 would be j02(f)(kappa_0,kappa_1), and you map this to [f]_{mu_0xnu} for an isomorphism. For your case, where mu_1 is NOT in the range of j_01, then I believe the answer is Yes, but this is more complicated, and I will post an answer later. –  Joel David Hamkins Feb 3 '10 at 13:38
    
I have posted a revised answer, for the case mu_1 not in ran(j_01). Here, the answer is affirmative. –  Joel David Hamkins Feb 11 '10 at 1:54
    
Thank you for your astonishing solution. –  Ant emyy Lee Feb 11 '10 at 6:53

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