Either {a1,a2,...an} must be the n nth roots of 1, or it must be the n-1 n-1th roots of 1 and 0.
So the polynomial is either x^n-1 or x*(x^n-1-1)
Here is a proof of alunw's claim:
Clearly all of the nonzero must have norm or else are all distinct. Hence for these we have . Moreover . Denoting the unit circle in the complex plane by , define
where is the class of . Then is an isomorphism . The nonzero are a finite subgroup of , which maps to a finite subgroup of , which is cyclic with generator with . This implies in turn that is cyclic with generator . Hence either or .
I think this can be done more simply. Not with any change in the central ideas, but without appealing to isomorphisms and equivalence classes.
Denote by the set without zero (the tex set minus notation is looking bad to me for some reason). Let . Assume without loss of generality are nonzero. If then since is closed and finite. Otherwise, is a subgroup for the same reason.
Lemma: If such that , then
Easy Proof: Since is a subgroup, and since , for all , so . Since the fundamental theorem of algebra guarantees at most distinct roots, and distinct roots exist, the result follows.
Then the claim follows because if , then the polynomial's roots must be the nth roots of unity and so the polynomial must be . Otherwise, consists of the (n-1)th roots of unity, so with zero we obtain .