1

Let $\xi=e^{ \frac{2 \pi i}{n}}$. Prove the following:

i) $$x^{n}-1= (x-1)(x- \xi)....(x- \xi^{n-1}).$$ And if $n$ is odd, then $$x^{n}+1= (x+1)(x+\xi)....(x+\xi^{n-1}).$$

ii) For $a$ and $b$ numbers, prove that

$$a^n-b^b=(a-b)(a- \xi b)....(a-\xi^{n-1}b),$$ and if $n$ is odd, then

$$a^n+b^b=(a+b)(a+\xi b)....(a+\xi^{n-1}b).$$

For $i$ I will use the following result in Rotman´s book:

Let $k$ be any field, perhaps finite. If $f (x), g(x) ∈ k[x]$, if $deg( f ) ≤ deg(g) ≤ n$, and if $f (a) = g(a)$ for $n + 1$ elements $a ∈ k$, then $f (x) = g(x).$

So in order to prove the first equality in (i) I must show that $x^{n}-1$ and $(x-1)(x- \xi)....(x- \xi^{n-1})$ share $n+1$ values. By other exercise is not hard to show that $z=1, z= \rho, z= \rho ^2,...,z= \rho^{n-1}$ are all the different solutions of $z^n=1$. So this gives me $n$ different shared values for both equations, but Rotman's result needs one more shared value to work. How I can find this other value?

For (ii) I'm not sure how to attack this problem; the only hint I got is to set $x = a/b$ if $b \neq 0$. But I do not really know how this helps. Any help with these two proofs will be appreciated. Thank you.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Cos
  • 2,097

1 Answers1

2

To prove $x^{n}-1= (x-1)(x- \xi)....(x- \xi^{n-1})$, note that both sides are polynomials of degree $n,$ and they both have the same value (viz., $0$) at the $n$ roots of unity and at $0$ (viz., $-1$).

Setting $x=\frac a b, \;b\ne0$, yields $\left(\frac a b \right)^n - 1 =\left (\frac a b -1\right)\left(\frac a b - \xi\right)...\left(\frac a b - \xi^{n-1}\right),$

and multiplying both sides by $b^n$ yields the desired result for (ii).

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
  • When $b=0$, we can't set $x=\frac a b$, but in that case the desired equality is simply $a^n=a^n$ – J. W. Tanner Mar 06 '19 at 02:26
  • When $x=0, (x-1)(x-\xi)...(x-\xi^{n-1})=-1(-\xi)...(-\xi^{n-1}) = -1(-\xi)^{1+...+n-1}=-1(-1)^{n-1}\xi^{n(n-1)/2}.$ When $n$ is even, this is $-1(-1)(\xi^{n/2})^{n-1}=(-1)^{n-1}=-1$, and when $n$ is odd, this is $-1(-1)^{n-1}(\xi ^n)^{(n-1)/2}=-1(1)(1)^{(n-1)/2}=-1$ – J. W. Tanner Mar 06 '19 at 02:35
  • I assumed OP made a typo. and the desired result should have $a^n-b^\mathbf n$, not $a^n-b^\mathbf b$ – J. W. Tanner Mar 06 '19 at 02:39