Fibonacci coefficient ring and its exact mod-31 specialization

The number $r$ is a primitive tenth root of unity, so its minimal polynomial is ${\mathrm{\Phi }}_{10}(x)=x^4-x^3+x^2-x+1$ and $[\mathbb{Q}(r):\mathbb{Q}]=4$. Its maximal real subfield is $\mathbb{Q}(\sqrt{5})$, and the nontrivial conjugate of $\varphi$ is $(1-\sqrt{5})\mathrm{/}2=-{\varphi }^{-1}$; hence the field norm of $\varphi$ from $\mathbb{Q}(\sqrt{5})$ to $\mathbb{Q}$ is $-1$. Therefore $\varphi$ is not a square in that real subfield, since the norm of a square would be a nonnegative rational square.

If $u\in \mathbb{Q}(r)$ satisfied $u^2=\varphi$, complex conjugation would give ${\bar{u}}^2=\varphi$, hence $\bar{u}=u$ or $\bar{u}=-u$. In the first case $u\in \mathbb{Q}(\sqrt{5})$, contradicting step 1.1. In the second case $u$ is purely imaginary in the chosen embedding, so $u^2$ is nonpositive real, contrary to $\varphi >0$. Thus $Y^2-\varphi$ is irreducible over $\mathbb{Q}(r)$ and $[\mathbb{Q}(r,s):\mathbb{Q}]=8$. The displayed monic relations therefore give the full degree-eight presentation; no additional relation is being assumed.

In ${\mathbb{F}}_{31}$, ${15}^{-1}=29$, ${\mathrm{\Phi }}_{10}(15)=0$, $1-15-29=19$, and $9^2=19$. Thus the assignments satisfy both defining relations and extend to the stated ring homomorphism. The images $15$ and $19$ are nonzero, so $r,\varphi$ and their powers are units; all denominators in the path projectors and normalized braid matrices are therefore invertible after specialization.

In the number field, complex conjugation fixes $\varphi ,s$ and sends $r$ to $r^{-1}$. Therefore a coefficient involving conjugation is first rewritten with $r^{-1}$ and only then reduced by $r^{-1}\mapsto 29$. No conjugation operation on ${\mathbb{F}}_{31}$ is required.