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Fact 1: If $M$ and $N$ are transitive models of ZF with the same ordinals, and $M \prec N$, then $M = N$.

Fact 2: If $M$ and $N$ are transitive models of ZFC with the same ordinals, and $j: M \to N$ is an elementary embedding that does not move ordinals, then $j$ is the identity map and $M=N$.

Question: Is it consistent that $M$ and $N$ are transitive models of ZF with the same ordinals, $j: M \to N$ is an elementary embedding that does not move ordinals, and $M \not= N$? Perhaps with $j(\mathbb{R}^M) \not= \mathbb{R}^M$?

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up vote 19 down vote accepted

Yes, this is possible. In fact, this is equiconsistent with ZF. For example, let ${\mathbb P}={\rm Add}(\omega,\omega_1)$ be the forcing that adds $\omega_1$ Cohen reals with finite conditions. Note that the definition of ${\mathbb P}$ is absolute between (transitive) models that agree on the value of $\omega_1$.

By standard arguments, any ${\mathbb P}$-generic $G$ is isomorphic to a product $G_0 \times G_1$, where $G_0$ is ${\mathbb P}$-generic over $V$ and $G_1$ is ${\mathbb P}$-generic over $V[G_0]$. Similarly, $G$ can be decomposed as $G'\times G''$ where $G'$ adds countably many Cohen reals and $G''$ is ${\mathbb P}$ generic over $V[G']$.

Let ${\mathbb R}_0 = {\mathbb R}^{V [G_0]}$ and ${\mathbb R}_1 = {\mathbb R}^{V [G]}$. In $V [G]$ we can define a nontrivial $j : L({\mathbb R}_0)\to L({\mathbb R}_1)$ that is the identity on the ordinals. The point is that any element of $L({\mathbb R})$ is definable from an ordinal and a real, so we can see it as a "term" $\tau(x,y)$ in an appropriate language, interpreted in $L({\mathbb R})$, when $x=\alpha$ and $y=r$ are an ordinal and a real, respectively.

[We can avoid talking about terms, of course, but it is perhaps a bit easier to see what follows if we insist in mentioning them.]

The fact above tells us that we have not much choice on how to define $j$, namely, we must have $$ j(\tau^{L({\mathbb R}_0)}(\alpha,r))=\tau^{L({\mathbb R}_1)}(\alpha,r) $$ for any term $\tau$, any ordinal $\alpha$, and any real $r\in{\mathbb R}_0$.

To see that $j$ works, note that there is a recursive way to assign to each formula $\varphi(\vec x,\vec y)$ a formula $\psi_\varphi(\vec x,\vec y)$ in such a way that (provably in ZF), for any ${\mathbb P}$-generic $H$, in $V[H]$ we have that $L({\mathbb R})\models\varphi(\vec \alpha,\vec r)$ iff $V[H]\models \psi_\varphi(\vec \alpha,\vec r)$ for any finite sequence of ordinals $\vec\alpha$ and any finite sequence of reals $\vec r$.

Now, and this is the key, there is a countable subset $X\subset\omega_1$ such that $r\in V [G_0\upharpoonright X]$, and we can find ${\mathbb P}$-generics $G'$ over $V[G_0\upharpoonright X]$ and $G^*$ over $V[G_0\upharpoonright X][G']$ such that $V[G_0\upharpoonright X][G'] = V [G_0]$ and $V[G_0\upharpoonright X][G^*]=V [G]$.

But then $L({\mathbb R}_0)\models \varphi(\vec \alpha, r)$ iff $V [G_0] \models\psi_\varphi(\vec\alpha,r)$ iff there is a $p\in G'$ such that in $V[G_0\upharpoonright X]$, $p$ forces that $\varphi(\vec\alpha,\vec r)$ holds, iff the empty condition forces it (this is homogeneity of ${\mathbb P}$, and is really the point of the whole thing) iff there is a $q\in G^*$ that forces it iff $V[G]\models\psi_\varphi(\vec\alpha,r)$ iff $L({\mathbb R}_1)\models\varphi(\vec \alpha,r)$.

Of course, it follows immediately from this that $j$ is well-defined, elementary, and the identity on ordinals.

The argument comes more or less verbatim from my thesis, but it is probably folklore. Certainly, A. Miller has very similar arguments, which makes me suspect he knew this ages before I asked myself the question. One can also use random forcing or several other forcing notions with minor variations.

One reason why I like this argument is because it gives us a cute way of proving that choice fails in $L({\mathbb R})$ after adding $\omega_1$ Cohen reals. On the other hand, adding countably many Cohen reals to $L$ is the same as only adding one, and the resulting $L({\mathbb R})$ is just $L[r]$ for some real $r$, so choice holds there.

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Very interesting, thanks! – Monroe Eskew Nov 22 '11 at 3:30

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