# Thread: Algebra, Problems For Fun (32)

1. ## 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.

2. 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)

3. 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 $p(x)=x\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^{r+1}-x$.

4. 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
$
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 $p(x)=x^n-x$.