TITLE: Show that there are non-well-founded models of Zermelo Fraenkel set theory
QUESTION [9 upvotes]: I have been working on this problem for several hours, and my understanding just isn't there. Here's what I've gathered:
Using downward Lowenheim-Skolem theorem, we know that any consistent set of sentences has a countable model, and for the sake of this problem we are assuming that ZF is indeed consistent.
The axiom of foundation in ZF seems to eliminate the possibility that there is a non-well-founded model, because it says that there is no infinitely descending membership sequence.
I believe I may need to employ compactness in the approach. Not sure what I need to proceed. TIA

REPLY [3 votes]: You can get this directly from the incompleteness theorem, without compactness. Let $M$ be a model of $\text{ZFC} + \lnot\text{Con}(\text{ZFC})$. 
Then:
1) $\omega^M$ is a model of Peano arithmetic, because ZFC proves that $\omega$ satisfies each axioms of Peano arithmetic.
2) $\omega^M$ is not the standard model of Peano arithmetic, because $\omega^M$ satisfies $\lnot\text{Con}(\text{ZFC})$, and the standard model does not.
3) $\omega^M$ is not well founded, because no nonstandard model of Peano arithmetic is well founded. If $x$ is a nonstandard element, then in particular it is a successor, and it follows that there is an infinite decreasing sequence of predecessors beginning at $x$.