Let $f(T) = \sum a_n T^n \in \mathbf{F}_p [[ T ]]$ be a power series. We'll say that the coefficients of $f(T)$ are equidistributed modulo $p$ if for every residue class $a$ modulo $p$, we have $$ \lim_{X \to \infty} \dfrac{1}{X} \cdot \# \{n<X : a_n \equiv a \mod p \} = \dfrac{1}{p}.$$
Suppose that $f(T), g(T) \in \mathbf{F}_p [[ T ]]$ are power series whose coefficients are equidistributed modulo $p$ in the above sense. Is it true that the coefficients of the product $f(T) \cdot g(T)$ are also equidistributed modulo $p$?

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    $\begingroup$ Taking $g(T)=f(-T)$, the product will be biased towards $0$. $\endgroup$ Aug 21, 2023 at 23:18

1 Answer 1


No. Let $f(T)$ be chosen uniformly at random from all power series with constant term nonzero and let $g(T) = 1/ f(T) $. Then $g(T)$ is also chosen uniformly at random from all power series with constant term nonzero. (Here uniformly means each possible sequence of first $n$ coefficients has equal probability, and the proof is just that the first $n$ coefficients of $f$ determine the first $n$ coefficients of $g$ and vice versa.)

So $f(T)$ and $g(T)$ are each equidistributed with probability $1$ but $f(T)g(T)=1$ is not equidistributed.

The same argument shows that any condition on $f$ and $g$ that is "generic" in the sense of holding for a random power series is not sufficient.

  • $\begingroup$ I see - are there conditions on $f(T)$ and $g(T)$ that would ensure that $f(T) \cdot g(T)$ is equidistributed? Perhaps some kind of independence condition that excludes the possibility of $g$ being determined by $f$ as in your counterexample? $\endgroup$ Aug 21, 2023 at 23:22
  • $\begingroup$ Since most sequences of elements of a finite field are equidistributed, I expect that most products of uniformly distributed power series will be uniformly distributed. $\endgroup$ Aug 21, 2023 at 23:47

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