Timeline for Laurent polynomials of the form $p(x)\cdot p(x^{-1})$

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Jun 26, 2021 at 12:24	comment	added	Duchamp Gérard H. E.		@Adam Oh, you are right. I conjecture that the non-zero polynomials of the form $𝑞(𝑥)=𝑝(𝑥)𝑝(𝑥^{−1})$ read $𝑞(𝑥)=(𝑥+𝑥^{−1})^𝑘𝑟(𝑥)$, $𝑟(𝑥)$ 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 $z↦1/z$. I must think about a precise statement. Thanks for interaction.
Jun 25, 2021 at 17:28	comment	added	Adam		$(x-i)(x^{-1}-i)$ 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.
Jun 25, 2021 at 9:48	comment	added	Chris Wuthrich		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.
Jun 25, 2021 at 6:32	history	edited	Adam	CC BY-SA 4.0	deleted 1 character in body
Jun 25, 2021 at 6:09	history	asked	Adam	CC BY-SA 4.0