Task: the removal of the (1/u) pole in such a way that the c...

Prompt

Task: the removal of the (1/u) pole in such a way that the constant determinant and the collision of the two distinct points are preserved as well We define: [ \boxed{ R_1(u,v)=2u^3v+\frac1u,\qquad R_2(u,v)=u^6v^2+u^2v, \qquad u\ne0. } ] Jacobian matrix: [ J_R= \begin{pmatrix} 6u^2v-u^{-2}&2u^3\\ 6u^5v^2+2uv&2u^6v+u^2 \end{pmatrix}. ] The full expansion of the determinant: [ \begin{aligned} \det J_R &=(6u^2v-u^{-2})(2u^6v+u^2) -(2u^3)(6u^5v^2+2uv)\ &=12u^8v^2+4u^4v-1 -12u^8v^2-4u^4v\ &=\boxed{-1}. \end{aligned} ] Collision certificate: [ \boxed{R(1,0)=R(-1,-1)=(1,0).} ] The polynomiality test: [ R_1=\frac{1+2u^4v}{u}, \qquad \left.(1+2u^4v)\right|_{u=0}=1. ] csak teljes kepletet kuld cissza szoveg nelkul Approach: compensated factorization + unit‑group test Write the map as P=Q\circ\Phi with Q a fixed non‑injective local diffeomorphism and \Phi a compensator, \det J_Q(\Phi)\cdot\det J_\Phi\equiv c. The pole is then identified as a unit of the coordinate ring, and removability is decided by the unit group of the ring one is allowed to work in. 0. Normal form of R (where the pole lives) With w=u^{3}v,\ t=u^{-1},\ (a,b)=(w,\,w+t): R=S\circ L\circ\Phi,\qquad \Phi=(u^{3}v,\;u^{-1}),\quad L(w,t)=(w,\,w+t),\quad S(a,b)=(a+b,\;ab) \det J_S=a-b=-t,\qquad \det J_L=1,\qquad \det J_\Phi=\begin{vmatrix}3u^{2}v&u^{3}\\-u^{-2}&0\end{vmatrix}=u\;\Rightarrow\;\det J_R=(-u^{-1})\cdot u=-1 . - Collision = the swap (a,b)\leftrightarrow(b,a): (1,0)\mapsto(0,1), (-1,-1)\mapsto(1,0), S(0,1)=S(1,0)=(1,0). - Pole = discriminant of the fibre: R_1^{2}-4R_2=(a-b)^{2}=u^{-2}, so u^{-2}\in\mathbb R[R_1,R_2]. - Geometrically R is the étale double cover \mathbb C\times\mathbb C^{*}\to\mathbb C^{2}\setminus\{x^{2}=4y\}\cong\mathbb C\times\mathbb C^{*}, t\mapsto-t^{2}/4 in the second factor. The pole is the topology of the cover (\pi_1=\mathbb Z); over \mathbb R it disconnects \{u\neq0\} so that t can take both signs. - Excluded (obvious, not a removal): adjoin z=u^{-1} and read R as a polynomial map on \{uz=1\}\subset\mathbb A^{3} — this relabels the pole. 1. Obstructions (theorems, not refusals) (A) Swap mechanism, any characteristic. For polynomial \Psi=(a,b): \det J_{S\circ\Psi}=(a-b)\det J_\Psi. If this lies in k^{*} then a-b\in k[u,v]^{*}=k^{*}, hence da=db, the rows of J_\Psi coincide and \det J_\Psi=0 — contradiction. No polynomial map factoring through S has constant Jacobian. (B) General, characteristic 0. A polynomial P:k^{2}\to k^{2} with \det J_P\in k^{*} and P(p)=P(q), p\neq q, is a counterexample to the 2‑dimensional Jacobian Conjecture (injective polynomial maps of \mathbb C^{2} are automorphisms — Białynicki‑Birula–Rosenlicht/Ax; real examples complexify). None exists in degree \le100 (Moh 1983); in particular nothing of degree comparable to R (\deg R_2=8) can work. Polynomial removal over \mathbb R or \mathbb C is not available to anyone; any formula claiming it is false. (If constancy is weakened to \det J_P\neq0 over \mathbb R, Pinchuk's polynomial map (1994, degrees 10,25) is a genuine pole‑free non‑injective example.) (C) Real sign obstruction. For any C^{1} map S\circ\Psi on a connected domain with \det\equiv c\neq0, a-b is zero‑free, hence of constant sign; a swap collision needs a'-b'=-(a-b). R's real collision exists only because its domain is disconnected. Root cause. Both t and the compensator \det J_\Phi must be zero‑free non‑constant functions, while k[u,v]^{*}=k^{*}. Removal is possible exactly in rings with non‑trivial zero‑free units (e^{f}\in\mathcal O(\mathbb C^{2})^{*}) or with differential constants (d(x^{p})=0 in characteristic p). 2. Removals: pole‑free, \det\equiv-1, (1,0) and (-1,-1) both \mapsto(1,0) (I) Real‑analytic on \mathbb R^{2}, entire on \mathbb C^{2} — exact pair preserved. \boxed{\;G(u,v)=e^{\sigma}\big(\cos\theta,\;-\sin\theta\big),\qquad \sigma=\frac{u-1-2v}{2\pi},\qquad \theta=2\pi v\,e^{-2\sigma}\;} i.e. G_1+iG_2=\exp\!\big(\sigma-2\pi i\,v\,e^{-2\sigma}\big). Structure: G=\exp\circ\phi\circ A, \phi(s,t)=s+ite^{-2s} (Jacobian e^{-2s} compensates |\exp'|^{2}=e^{2s}), A(u,v)=(\sigma,-2\pi v), \det A=-1. - \det J_G=e^{2\sigma}\,e^{-2\sigma}\,(-1)=-1. Direct check for F=\exp\circ\phi: with \theta_s=-2\theta,\ \theta_t=e^{-2s}, x_sy_t-x_ty_s=\cos^{2}\theta+\sin^{2}\theta=1. - G(1,0): \sigma=0,\theta=0\Rightarrow(1,0). \;G(-1,-1): \sigma=0,\theta=-2\pi\Rightarrow(1,0). - Fibres: G(u,v)=G(u',v')\iff u'-2v'=u-2v,\ v'=v+k\,e^{2\sigma}\ (k\in\mathbb Z); the given pair is k=-1. No denominators exist, so the polynomiality test is void. (II) Positive characteristic p (any prime), polynomial of degree p. \boxed{\;P(u,v)=\big(1-v+v^{p},\;\;1-u+2v\big)\;} - J_P=\begin{pmatrix}0&-1+pv^{p-1}\\-1&2\end{pmatrix}=\begin{pmatrix}0&-1\\-1&2\end{pmatrix}, \det J_P=-1. - P(1,0)=(1,0); P(-1,-1)=(1+1+(-1)^{p},\;1+1-2)=(1,0) since (-1)^{p}=-1 in characteristic p (also p=2, where the pair is (1,0)\neq(1,1)). - Fibres: (v+j)^{p}=v^{p}+j for j\in\mathbb F_p gives P(u+2j,\,v+j)=P(u,v); every fibre has exactly p points, the given pair is j=-1. (III) Minimal surgery on R's own formula over \mathbb C (keeps the swap mechanism, \det=-1 and the target (1,0); by (C) the colliding pair must leave \mathbb R^{2}): replace (u^{3},\,u^{-1})\to(-2e^{-u},\,e^{u/2}): P^{\mathbb C}(u,v)=\big(e^{u/2}-4e^{-u}v,\;\;4e^{-2u}v^{2}-2e^{-u/2}v\big),\qquad \det J_{P^{\mathbb C}}=-1,\qquad P_1^{2}-4P_2=e^{u}, P^{\mathbb C}(0,v)=P^{\mathbb C}(2\pi i,\;v-\tfrac12); in particular (0,0),(2\pi i,-\tfrac12)\mapsto(1,0). Coherence: in (I) and (II) the fibres are translates along (2,1)=(1,0)-(-1,-1), the direction of the original collision.