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Recall Kunen's result: If $j:V \to M$ is a nontrivial elementary embedding then $V \ne M$. Assume $AD$ is true in $L(\mathbb{R})$. Let $j : L(\mathbb{R}) \to M$ be a nontrivial elementary embedding. Must we have $L(\mathbb{R}) \ne M$? Thanks.

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I think it is still open whether or not the theorem holds without choice. In particular in AD models. – Asaf Karagila Sep 14 '12 at 23:06
I also think you mean $L(\mathbb R)$, which is a choiceless (possibly) model. Whereas in modern notation the $[]$ brackets always produce a model of ZFC. – Asaf Karagila Sep 14 '12 at 23:13
Let me edit that. – user33714 Sep 14 '12 at 23:15
up vote 2 down vote accepted

The existence of a nontrivial elementary embedding $j: L(\mathbb{R}) \to L(\mathbb{R})$, or equivalently, the existence of $\mathbb{R}^\sharp$, is consistent with the statement that AD holds in $L(\mathbb{R})$. Both are consequences of the existence of large cardinals, such as $\omega$ many Woodin cardinals with a measurable cardinal above them. Also, both are conseqences of stronger determinacy axioms, for example AD$_\mathbb{R}$, the axiom of determinacy for games where the moves are reals rather than integers.

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Kunen result was that if $V$ is a model of ZFC then there is no surjective elementary embedding which is not the identity.

Your question asks about AD, but even if we just require the axiom of choice to fail -- it is unknown whether or not this can be done. There are some very good reasons to believe that this should hold, because the proof of Kunen's theorem uses choice quite extensively, however things are unclear so far.

But... your question is slightly vague in the sense that it is unclear whether or not you want this embedding to be definable internally or externally (with parameters, naturally). For example if $0^\#$ exists then there is a nontrivial elementary embedding from $L$ to itself, but of course this embedding is definable in $V$ but not in $L$ (of course $V\neq L$ follows from the existence of a sharp).

Whether or not this would remain true in $L(\mathbb R)$ I cannot say, it is possible that if we assume AD there are enough large cardinals to define in $V$ an elementary embedding, but that would be external to $L(\mathbb R)$, which is an important point unclear in your question.

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