A question related to cyclotomic polynomials in -x

Question

Consider the following problem from Hungerford Algebra :

If n is odd, then $g_{2n}(x) = g_n(-x)$ where $g_n(x) $ are cyclotomic polynomials over $\mathbb{Q}$.

So, $g_{2n} (x) = \frac {x^{2n}-1} {\prod_{d|2n, d<2n} g_d(x) }$ and $g_n(-x) = \frac{(-1)(-x)^n -1}{x+1 ...} $ can be written similarly . The problem I am facing is that in denominator of g_n(-x) , the $g_d(x)$ terms will be have difference due to $(-x)^n$ in the $g_d(x)$, so I am not able to get an idea on how should I proceed.

So, Can you please help me?

Answer 1

Some hints:

• Irreducible polynomials over $\Bbb Q$ are separable. In particular, division by such a polynomial amounts to evaluating at its roots and checking they are zero.

If $\xi$ is a root of $g_{n}(-x)$ then $(-\xi)^{n} = (-1)^n\xi^n = 1$ and so $\xi^{2n} = 1$. Hence $g_n(-x) \mid g_{2n}(x)$. But why are these two equal?

Since $n$ is odd we have $\deg g_{2n} = \varphi(2n) = \varphi(n)$, and both are monic, so they must coincide.

Answer 2

Here is a way that involves essentially no computation at all. We note that $x^{2n}-1=(x^n-1)(x^n+1)$. Since $x^n-1$ is the product of all cyclotomic polynomials $g_m$ for $m\mid n$, $x^n+1$ must be the product of all cyclotomic polynomials $g_m$ for $m\mid 2n$ and $m\nmid n$.

Now let $n$ be odd. Then $x^n+1=-((-x)^n-1)$, so has the same factorization into irreducibles, but with $-x$ instead of $x$. Furthermore, the set of $m$ such that $m\mid 2n$ and $m\nmid n$ is simply $2k$ for $k\mid n$. Thus the map $i\mapsto 2i$ induces a map from the irreducible factors of $x^n-1$ to those of $x^n+1$. We just have to decide whether it maps $g_i$ to $g_{2i}$. But certainly this is true for $n=1$, and now by induction the result is true, since for any given $n$ there is a single factor for which the result is not already known by the inductive hypothesis.