Assume one has. for $V$ and for some transitive class $M$, an elementary embedding
$j$: $V$$\rightarrow$$M$ and that $j$$\neq$$id$, where $id$ is the identity.
If $V$ and $M$ satisfy $ZFC$ then the following Theorem holds
Thm. Let $\alpha$ be an ordinal.
(i) For every $\alpha$, $j$($\alpha$)$\ge$$\alpha$
(ii) $j$ moves some ordinal.
Let $\delta$ be the least ordinal moved by $j$. $\delta$ is called the critical point of $j$.
It can be proven in $ZFC$ that $\delta$ is always a cardinal.
Suppose now that $j$: $V$$\rightarrow$$M$, and $V$ and $M$ both satisfy $ZF$+$\lnot$$AC$. How can the critical point $\delta$ be defined when there exist incomparable cardinals?
I hope that this is not too silly a question. If it turns out to be silly, I will happily delete it.
(Addendum: Regarding Noah's answer to my question ("...even though the cardinalities of $V$ may not be well-ordered, the cardinals and ordinals definitely will be..."), in $ZF$+$\lnot$$AC$, there are cardinals which are not ordinals, and these may not be comparable. Since under a nontrivial elementary embedding no ordinal would be moved, could such cardinals be critical points of such an embedding?)
You can define the critical point in exactly the same way.
First, note that even though the cardinalities of $V$ may not be well-ordered, the cardinals and ordinals definitely will be. EDIT: BY "cardinality," I mean cardinality in the most general sense; by "cardinal," I mean cardinality of a well-orderable set, that is, "$\aleph$-number." Equivalently, by "cardinal" I mean "ordinal which is not in bijection with any smaller ordinal."
So we still have the result that $j(\alpha)$ is an ordinal $\ge\alpha$, for every ordinal $\alpha$.
Moreover, by transfinite induction (which doesn't require choice) if $j(\alpha)=\alpha$ for every ordinal $\alpha$ then $j$ is trivial - again, this doesn't use choice.
So the answer is, the critical point can be defined in exactly the same way - as the least ordinal moved by $j$.