4
$\begingroup$

Let $R$ be a commutative, associative ring with $1$ and let $\tau: R[x^{\pm 1}]\to R[x^{\pm 1}]$ be the $R$-algebra involution $\tau(x)=x^{-1}.$ Then it is natural to ask which elements of the invariant subalgebra $R[x^{\pm 1}]^\tau$ are of the form $p\cdot \tau(p)$ for some $p\in R[x^{\pm 1}]$.

This problem reduces to a question whether each $$a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$$ equals $p(x)p(x^{-1}),$ for some $p(x)=\sum_{i=0}^n c_ix^i$, which leads to the system of equations: $$\begin{cases} c_0^2+...+c_n^2=a_0\\ c_0c_1+...+c_{n-1}c_n=a_1\\ ...\\ c_0c_n=a_n. \end{cases} $$

I suspect that this problem has been studied already and I am hoping to be pointed in the right direction. I expect that this system have a solution for algebraically closed fields $R$ (at least when characteristic $\ne 2$), but don't know the proof. However, the most interesting case for me is $R=\mathbb Z$. I wonder if there is a manageable description of $a_0,...,a_n\in \mathbb Z$ for which the above system has a solution in integers.

Improve this question
$\endgroup$
3
  • $\begingroup$ Sounds like an Tate cohomology group $\hat H^0(\langle \tau\rangle, ?)$. The invariant subalgebra is what is fixed by $\tau$ and you are asking for the image of $1+\tau\in\mathbb{Z}[\langle\tau\rangle]$. Just not clear what multiplicative abelian group $?$ to take. $\endgroup$
    – Chris Wuthrich
    Commented Jun 25, 2021 at 9:48
  • $\begingroup$ $(x-i)(x^{-1}-i)$ does not have a constant term. Also, are you parametrizing $\tau$-invariant polynomials or polynomials $p(x)p(x^{-1})$? If the $\tau$-invariant ones, then how do you show that $1$ and $-1$ have even multiplicity? I think this is at the heart of showing that these two classes coincide over algebraically closed fields. $\endgroup$
    – Adam
    Commented Jun 25, 2021 at 17:28
  • $\begingroup$ @Adam Oh, you are right. I conjecture that the non-zero polynomials of the form $š‘ž(š‘„)=š‘(š‘„)š‘(š‘„^{āˆ’1})$ read $š‘ž(š‘„)=(š‘„+š‘„^{āˆ’1})^š‘˜š‘Ÿ(š‘„)$, $š‘Ÿ(š‘„)$ having a constant term and a multiset of roots $\mathcal{O}_r$ as I described i.e. (i) $±1$ each of even multiplicities (ii) $\mathcal{O}_r$ is closed by $zā†¦1/z$. I must think about a precise statement. Thanks for interaction. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 26, 2021 at 12:24

0

Reset to default

You must log in to answer this question.