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Let $SRMI_q(2n)$ denote the number of self-reciprocal irreducible monic polynomials of even degree $2n$ over the finite field $\mathbf{F}_q$ with $q$ elements. Recall that a polynomial $p(x) \in \mathbf{F}_q[x]$ of degree $n$ is self-reciprocal if $p(x)=x^np(x^{-1})$. According to Thm 3.1.20 of Handbook of Finite Fields by Mullen and Panario we have that \begin{equation*} SRMI_q(2n) = \frac{1}{2n}\sum_{\text{odd $d \mid n$}} \mu(d)(q^{n/d}-1), \qquad SRMI_q(2n) = \frac{1}{2n}\sum_{\text{odd $d \mid n$}} \mu(d)q^{n/d} \end{equation*} depending on whether $q$ is odd or even. The sums range over all odd divisors of $n$. Now comes my question. Is it possible to write the infinite product \begin{equation*} \prod_{n \geq 1} \frac{1}{(1-x^{2n})^{SRMI_q(2n)}} \end{equation*} as a rational function?

What I have in mind is an analogue of the well-known formula \begin{equation*} \prod_{n \geq 1} \frac{1}{(1-x^n)^{I_q(n)}} = \frac{1}{1-qx} \end{equation*} where $I_q(n)$ is the number of irreducible monic polynomials over $\mathbf{F}_q$.

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  • $\begingroup$ What is a self-reciprocal polynomial? $\endgroup$ – Zach Teitler Sep 21 '17 at 16:49
  • $\begingroup$ Please see edited version of my question. $\endgroup$ – Jesper M. Moller Sep 21 '17 at 17:20
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Let $A(x)=1+xB(x) \in \mathbb{C}[[x]]$ be a power series. It is rational iff $A(x) = \frac{\prod (1-\alpha_i x)}{\prod (1-\beta_j x)}$ for some $\alpha_i,\beta_j \in \mathbb{C}$. A standard calculation shows that $$A(x) = \frac{\prod (1-\alpha_i x)}{\prod (1-\beta_j x)} \leftrightarrow x \frac{A'(x)}{A(x)} = \sum_{n \ge 1} (\sum a_i^n - \sum b_j^n)x^n.$$ If we write $A(x)$ as an infinite product $$A(x) = \prod_{n \ge 1} (1-x^n)^{-a_n}$$ for some $a_n \in \mathbb{C}$ (this is always possible, in a unique way), we have $$x \frac{A'(x)}{A(x)} = \sum_{n \ge 1} (\sum_{d \mid n} a_d d)x^n.$$

Thus, what we really need to understand is $\sum_{d \mid n} a_d d$ rather than $a_n$. In our case, $a_{2n} = SRMI_q(2n), a_{2n-1} = 0$. I will focus on the even $q$ case, for simplicity. Let $n$ be a positive integer. We have, by changing the order of summation,

$$\sum_{d \mid n} a_d d =\sum_{2d \mid n} a_{2d} 2d=\sum_{m \mid n/2} q^m \sum_{i \mid n/(2m), \text{ odd}} \mu(i).$$ The identity $\sum_{i \mid s}\mu(i)=1_{s=1}$ implies that $\sum_{i \mid n/(2m), \text{ odd}} \mu(i) = 1_{n/{2m} \text{ is a power of 2}}$, and thus, if we let $2^{v_2(n)}$ be the highest power of $2$ dividing $n$, then $$(*) \sum_{d \mid n} a_d d=\sum_{i=0}^{v_2(n)-1} (q^{2^{-(i+1)}})^n.$$

It is not hard to see that the sum $(*)$ cannot be of the form $\sum a_i^n - \sum b_j^n$. (Sketch of Proof: The values of $\alpha_i,\beta_j$ are the poles of $x\frac{A'(x)}{A(x)}$, while $(*)$ implies that $x\frac{A'(x)}{A(x)}$ has infinitely many poles at $q^{2^{-(i+1)}}$.)

[Addendum] Identity $(*)$ also implies that $$A(x) = \prod_{i \ge 1}(1-qx^{2^i})^{-2^{-i}}$$ which leads to the functional equation $$\frac{A(x^2)}{A^2(x)} = 1-qx^2.$$ As for the odd $q$ case, the same arguments lead to $$A(x) = \prod_{i \ge 1}(\frac{1-qx^{2^i}}{1-x^{2^i}})^{-2^{-i}}, \frac{A(x^2)}{A^2(x)} = \frac{1-qx^2}{1-x^2}.$$

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A complement to Ofir's answer: Check out the paper:

Meyn, Helmut; Götz, Werner, Self-reciprocal polynomials over finite fields, Sémin. Lothar. Comb. 21, B21d, 8 p. (1989). ZBL1011.11507.

You will note that your counting is essentially equivalent to counting irreducible polynomials with the coefficient of $x$ equal to $1.$ (essentially, because it does not account for the stray irreducible reciprocal polynomial $x+1.$)

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This is just a few comments

Why are you excluding the odd degree self-reciprocal polynomials?

Is there a formula for $$ \prod_{n \geq 1} \frac{1}{(1-x^{2n})^{I_q(2n)}} ?$$

For any of these variations you could multiply out a few terms and see if a pattern emerges. If it is anything like what you want to emulate, it should be clear.

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  • $\begingroup$ Because odd-degree reciprocal polynomials (with the exception of $x+1$) are reducible, being divisible by $x+1.$ $\endgroup$ – Igor Rivin Sep 21 '17 at 18:34

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