Comparing countable models of ZFC Let us consider the class $\cal C$ of countable models of ZFC. For ${\mathfrak A}=(A,{\in}_A)$ and ${\mathfrak B}=(B,{\in}_B)$ in $\cal C$ I say that ${\mathfrak A}<{\mathfrak B}$ iff there is a injective map $i: A \to B$ such that $x {\in}_A y \Leftrightarrow i(x) {\in}_B i(y)$ (note that this is a much weaker requirement for $i$ than to be an elementary embedding). My two questions are :
(1) Is there a simple construction of two incomparable models ${\mathfrak A},{\mathfrak B}$ ?
(i.e. neither ${\mathfrak A}<{\mathfrak B}$ nor ${\mathfrak B}<{\mathfrak A}$).
(2) Given two models ${\mathfrak A},{\mathfrak B}$ in $\cal C$, is there always a third model ${\mathfrak C}$ in $\cal C$ such that ${\mathfrak A}<{\mathfrak C}$ and ${\mathfrak B}<{\mathfrak C}$ ?
 A: (2) seems true. Choose models with universes $\lbrace m_j\vert j<\omega\rbrace$, $\lbrace n_j\vert j<\omega\rbrace$ Consider language $L=\lbrace \in, a_i,b_i\vert i<\omega\rbrace$, and the theory $T=ZFC\cup\lbrace a_i\neq a_j\vert i,j\in\omega, m_i\neq m_j\rbrace\cup \lbrace a_i\in a_j\vert i,j\in\omega, m_i\in^{M_1} m_j\rbrace \cup \ldots$
It is clear that if $T$ is consistent, we can obtain the countable model in which $M_1,M_2$ embed monomorphically by downward Skolem.
Choose a finite fragment of $T$. The formulas of $T$ don't relate $a$ and $b$ in any way, so it is effectively a fragment of ZFC plus two finite (well-founded and consistent) membership+non-membership graphs. But any such graph can be realized by a finite set in any model of ZFC, so by compactness $T$ is consistent.
For (1) i think you can try to look at models which realize different subtrees of the Cantor tree (as subgraphs of their membership graphs). For example, one of them could have infinite descending sequence and other might not. It should be doable by omitting types theorem.
A: Concerning question (1).
I became very interested in this question last year---obsessed
with it, actually---when I found myself unable to prove that any
of the natural-seeming examples were actually instances of
incomparability (for example, none of the approaches suggested in the various comments actually work). After my numerous attacks on it failed, I began
seriously to doubt the strong intuition underlying the question,
that there should be incomparable models. Eventually, I was a able to show that indeed, any two countable models are comparable by embeddability. My paper is available at:


*

*J. D. Hamkins, "Every countable model of set theory
embeds into its own constructible universe", also at the math arχiv. 


The main theorems are:
Theorem 1. Every countable model of set theory $\langle
M,{\in^M}\rangle$ is isomorphic to a submodel of its own
constructible universe $\langle L^M,{\in^M}\rangle$. Thus, there
is an embedding
 $$j:\langle M,{\in^M}\rangle\to \langle L^M,{\in^M}\rangle$$
that is elementary for quantifier-free assertions in the language
of set theory.
The proof uses universal digraph combinatorics, including an
acyclic version of the countable random digraph, which I call the
countable random $\mathbb{Q}$-graded digraph, and higher analogues
arising as uncountable Fraisse limits, leading eventually to what
I call the hypnagogic digraph, a set-homogeneous, class-universal,
surreal-numbers-graded acyclic class digraph, which is closely
connected with the surreal numbers. The proof shows that $\langle
L^M,{\in^M}\rangle$ contains a submodel that is a universal
acyclic digraph of rank $\text{Ord}^M$, and so in fact this model
is universal for all countable acyclic binary relations of this
rank. When $M$ is ill-founded, this includes all acyclic binary
relations.
The method of proof also establishes the following, which answers
question (1). Version 2 on the archive, which will become visible
in a few days, cites this question and Ewan Delanoy.
Theorem 2. The countable models of set theory are linearly
pre-ordered by embeddability: for any two countable models of set
theory $\langle M,{\in^M}\rangle$ and $\langle N,{\in^N}\rangle$,
either $M$ is isomorphic to a submodel of $N$ or conversely.
Indeed, the countable models of set theory are pre-well-ordered by
embeddability in order type exactly $\omega_1+1$.
The proof shows that the embedability relation on the models of
set theory conforms with their ordinal heights, in that any two
models with the same ordinals are bi-embeddable; any shorter model
embeds into any taller model; and the ill-founded models are all
bi-embeddable and universal.
The proof method arises most easily in finite set theory, showing
that the nonstandard hereditarily finite sets $\text{HF}^M$ coded
in any nonstandard model $M$ of PA or even of $I\Delta_0$ are
similarly universal for all acyclic binary relations. This
strengthens a classical theorem of Ressayre, while simplifying the
proof, replacing a partial saturation and resplendency argument
with a soft appeal to graph universality.
Theorem 3. If $M$ is any nonstandard model of PA, then
every countable model of set theory is isomorphic to a submodel of
the hereditarily finite sets $\langle \text{HF}^M,{\in^M}\rangle$
of $M$. Indeed, $\langle\text{HF}^M,{\in^M}\rangle$ is universal
for all countable acyclic binary relations.
In particular, every countable model of ZFC and even of ZFC plus
large cardinals arises as a submodel of
$\langle\text{HF}^M,{\in^M}\rangle$. Thus, inside any nonstandard
model of finite set theory, we may cast out some of the finite
sets and thereby arrive at a copy of any desired model of infinite
set theory, having infinite sets, uncountable sets or even large
cardinals of whatever type we like.
The article closes with a number of questions, which you may find
on my blog post about the article. I plan to make
some mathoverflow questions about them in the near future.
