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Math Help - Algebra, Problems For Fun (32)

  1. #1
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    Algebra, Problems For Fun (32)

    Given an integer n \geq 1, find all polynomials p(x)=(x-a_1)(x-a_2) \cdots (x-a_n) \in \mathbb{C}[x], where a_i \neq a_j if i \neq j and \{a_1,a_2, \cdots , a_n \} is closed under multiplication.
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  2. #2
    Member alunw's Avatar
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    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)
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Here is a proof of alunw's claim:

    Clearly all of the nonzero a_j must have norm 1 or else a_j, {a_j}^2,... are all distinct. Hence for these a_j we have a_j=\exp(b_j 2\pi i). Moreover a_ja_k=\exp((b_j+b_i)2\pi i). Denoting the unit circle in the complex plane by S_1, define

    f:S_1 \rightarrow \mathbb{R}/(2\pi\mathbb{R})

    \exp(b_j 2\pi i) \mapsto \overline{b_j}

    where \overline{b_j} is the class of b_j \: (\mbox{mod } 2\pi). Then f is an isomorphism S_1 \cong \mathbb{R}/(2\pi\mathbb{R})^+. The nonzero a_j are a finite subgroup H of S_1, which maps to a finite subgroup of \mathbb{R}/(2\pi\mathbb{R})^+, which is cyclic with generator \overline{2\pi/r} with r \in \mathbb{Z}. This implies in turn that H is cyclic with generator h=\exp( 2\pi i/r). Hence either p(x)=\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^r-1 or p(x)=x\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^{r+1}-x.
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  4. #4
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    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 X^* the set X without zero (the tex set minus notation is looking bad to me for some reason). Let S=\{a_1,\ldots,a_n\}. Assume without loss of generality a_1,\ldots, a_{n-1} are nonzero. If a_n\ne 0<br />
then S\leq\mathbb{C}^* since S is closed and finite. Otherwise, S^* is a subgroup for the same reason.

    Lemma: If  H\leq\mathbb{C}^* such that |H|=n, then H=\{\zeta\in\mathbb{C}^*:\zeta^n-1=0\}
    Easy Proof: Since H is a subgroup, 1\in H and since |H|=n, for all \zeta\in H, \zeta^n=1 so \zeta^n-1=0. Since the fundamental theorem of algebra guarantees at most n distinct roots, and n distinct roots exist, the result follows.

    Then the claim follows because if 0 \notin S, then the polynomial's roots must be the nth roots of unity and so the polynomial must be P(x)=x^n-1. Otherwise, S^* consists of the (n-1)th roots of unity, so with zero we obtain p(x)=x^n-x.
    Last edited by siclar; July 8th 2009 at 09:31 PM. Reason: to prevent horribly abusing notation
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