# Thread: Algebra, Problems For Fun (32)

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

Given an integer $\displaystyle n \geq 1,$ find all polynomials $\displaystyle p(x)=(x-a_1)(x-a_2) \cdots (x-a_n) \in \mathbb{C}[x],$ where $\displaystyle a_i \neq a_j$ if $\displaystyle i \neq j$ and $\displaystyle \{a_1,a_2, \cdots , a_n \}$ 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 $\displaystyle a_j$ must have norm $\displaystyle 1$ or else $\displaystyle a_j, {a_j}^2,...$ are all distinct. Hence for these $\displaystyle a_j$ we have $\displaystyle a_j=\exp(b_j 2\pi i)$. Moreover $\displaystyle a_ja_k=\exp((b_j+b_i)2\pi i)$. Denoting the unit circle in the complex plane by $\displaystyle S_1$, define

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

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

where $\displaystyle \overline{b_j}$ is the class of $\displaystyle b_j \: (\mbox{mod } 2\pi)$. Then $\displaystyle f$ is an isomorphism $\displaystyle S_1 \cong \mathbb{R}/(2\pi\mathbb{R})^+$. The nonzero $\displaystyle a_j$ are a finite subgroup $\displaystyle H$ of $\displaystyle S_1$, which maps to a finite subgroup of $\displaystyle \mathbb{R}/(2\pi\mathbb{R})^+$, which is cyclic with generator $\displaystyle \overline{2\pi/r}$ with $\displaystyle r \in \mathbb{Z}$. This implies in turn that $\displaystyle H$ is cyclic with generator $\displaystyle h=\exp( 2\pi i/r)$. Hence either $\displaystyle p(x)=\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^r-1$ or $\displaystyle p(x)=x\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^{r+1}-x$.

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 $\displaystyle X^*$ the set $\displaystyle X$ without zero (the tex set minus notation is looking bad to me for some reason). Let $\displaystyle S=\{a_1,\ldots,a_n\}$. Assume without loss of generality $\displaystyle a_1,\ldots, a_{n-1}$ are nonzero. If $\displaystyle a_n\ne 0$ then $\displaystyle S\leq\mathbb{C}^*$ since $\displaystyle S$ is closed and finite. Otherwise, $\displaystyle S^*$ is a subgroup for the same reason.

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

Then the claim follows because if $\displaystyle 0 \notin S$, then the polynomial's roots must be the nth roots of unity and so the polynomial must be $\displaystyle P(x)=x^n-1$. Otherwise, $\displaystyle S^*$ consists of the (n-1)th roots of unity, so with zero we obtain $\displaystyle p(x)=x^n-x$.