Properties of invariant polynomials

Question

Let $\alpha$ be a primitive $k$th root of unity. If $f(z)$ is a complex polynomial of degree $n$ and $f(\alpha^j z) = f(z) $ for all $j \in \mathbb{Z}$ what can we say about $f$?

This type of functions is called invariant functions. I couldn't find a textbook or resource about this topic.

So have you dealt with the case $k=2$? That ought to give you a good hint for the general case. — ancient mathematician, Feb 18 '22 at 15:52
If $\alpha\ne1$, I think the coefficient of each $z^m$ in $f$ must be zero whenever $m$ is not an integer multiple of $k$. That is, $f$ is a polynomial in $z^k$. — Ramen Nii-chan, Feb 18 '22 at 16:03
Hint: $;f(z)+f(\alpha z)+f(\alpha^2 z)+\dots+f(\alpha^{n-1} z)=\dots$ — dxiv, Feb 18 '22 at 16:25
@dxiv it is equal to $nf(z)$., but does that gice idea about the degree of $f$? — Adam_math, Feb 18 '22 at 17:02
@Adam_math I meant for an arbitrary polynomial $,f(z),$ in general, see here for example. — dxiv, Feb 18 '22 at 17:51

score 1 · Accepted Answer · edited Feb 19 '22 at 21:18

1

Write $f=f_0+\cdots +f_n$ in its homogeneous pieces.

$f(\alpha^j z)=f(z)$ implies that

$f_t(\alpha^j z)=\alpha^{jt}f_t= f_t$

and so $jt\equiv 0 \bmod k$.

Thus $z^j f= z^k F$

where $F=\sum_{\{t \ : \ jt\equiv 0 \bmod k\}}f_{\frac{jt}{k}}$

edited Feb 19 '22 at 21:18

user26857

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answered Feb 19 '22 at 19:52

Federico Fallucca

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Properties of invariant polynomials

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