Irreducible function on Q[z]

Let f(z) \in \mathbb{Q}[z] be irreducible with degree n>1. If f has a root on the unit circle then n is even and

    \[z^n f\left( \frac{1}{z} \right) = f(z)\]

Solution

Let \alpha be a root of f with |\alpha|=1. Since f has real coefficients \bar{\alpha}= \frac{1}{\alpha} is also a root of f. The product z^n f\left( \frac{1}{z} \right) is a polynomial in \mathbb{Q}[z] of degree n ( its leading coefficient is f(0) ) with root \alpha. By the irreducibility of f we have

(1)   \begin{equation*} z^n f\left( \frac{1}{z} \right) =  c f(z) \end{equation*}

for some non zero rational number c. Setting z=1 we have that f(1)=c f(1). Since f(1) \neq 0  , by our hypotheses , c=1 hence z^n f\left( \frac{1}{z} \right) = f(z) . Setting z=-1 we get that f(-1)=(-1)^n f(-1) and because f(-1) \neq 0 we deduce that n is even.

Note: The above tells us that f(z) can be expressed in terms of z + \frac{1}{z}.

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