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Theorem 1 and the all-finite template conversion are covered by the graded whole-conclusion review. The project argument is credited to Plasma AI. It supplies the template input of Proposition 6 of the [[research/leads/equilateral_two_scale/transcendental_scales|transcendental account]]. The field hypotheses of the other rigidity lemmas are checked in the [[research/leads/equilateral_two_scale/rigidity_extension|rigidity extension]], which composes this input with the triangle argument.

Let $E$ be the complex algebraic numbers, let $F=E\cap\mathbb R$, and let $x\in\mathbb R$ be transcendental. Write $L$ for the algebraic closure of $E(x)$ in $\mathbb C$, with its usual conjugation, and $K_L=\{u\in L:u\bar u=1\}$. All vector spaces and linear functionals in this account have scalars in $E$, unless a different field is indicated.

Theorem 1. There are no finite $Q=\{q_1,\ldots,q_J\}\subset K_L$ and $E$-linear $\lambda:L\to E$ such that $\lambda(1)=1$ and

$$\prod_{j=1}^J\lambda(q_j u)=0\qquad(u\in K_L).$$

Consequently every finite subset of $K_L$ has the template property of Proposition 6 of the transcendental account, by Proposition 8 there.

Reduction to a finite function field

Suppose such $Q$ and $\lambda$ exist. Choose a finite extension $N/E(x)$ inside $L$ containing $Q$ and stable under conjugation. Put $N_0=N\cap\mathbb R$, so $N=N_0(i)$ and $N_0/F(x)$ is finite. Define

$$T=\frac{x+i}{x-i},\qquad B=E(T)=E(x),\qquad d=[N:B], \qquad K_N=N\cap K_L.$$

Thus $T$ is a transcendental unit. The algebra $A=\operatorname{span}_E K_N\subset N$ contains $T,T^{-1}$ and every $q_j$, and each $q_j$ is a unit of $A$. It is stable under conjugation. We restrict $\lambda$ to $N$; the assumed cover still holds on $K_N$ and, by homogeneity, on every nonzero $E$-multiple of an element of $K_N$.

We need units in linear general position over $B$. There are arbitrarily large finite subsets of $K_N$ such that every $d$ distinct members are a $B$-basis of $N$. Here is the number-field general-position argument of Lemma 3.4 of the OpenAI plane-coloring manuscript, with the infinite base field $F(x)$ in place of $\mathbb Q$. For $z\in N_0$ use the unit $(1+iz)/(1-iz)$, and express $z$ in a basis of $N_0/F(x)$. The coordinates of this unit in $N/B$ are rational functions of those parameters, by the matrix formula for inversion in a finite extension. Their denominators are nonzero at every tuple in $F(x)^d$. No nonzero $B$-linear functional $\ell$ vanishes at all these units: for fixed $a\in N_0$, apply $\ell$ to $(1+iha)/(1-iha)$ for $h\in F(x)$ and expand the resulting rational identity at $h=0$. Its constant and linear coefficients give $\ell(1)=\ell(ia)=0$. As $N_0$ spans $N$ over $B$, this would give $\ell=0$. The product of the nonzero numerator polynomials for finitely many hyperplanes cannot vanish on the whole grid over the infinite field $F(x)$. Choosing each new unit outside the spans of all smaller subsets of the preceding choices proves the assertion.

Choose a $B$-basis $b_1=1,b_2,\ldots,b_d$ of units, and choose $n>J(d-1)$ units $s_1,\ldots,s_n$ in this general position. Write

$$q_j s_k=\sum_{\ell=1}^d r_{jk\ell}(T)b_\ell, \qquad r_{jk\ell}\in E(T).$$

For every fixed $j$, any $d$ of the row vectors $r_{jk}=(r_{jk\ell})_\ell$ form an invertible matrix over $E(T)$.

A rational Cauchy transform forced by the cover

For $a\in E$ define the vector and the scalar transforms

$$F_\lambda(a)=\left(\lambda\left(\frac{b_\ell}{T-a}\right)\right)_\ell, \qquad H_{jk}(a)=\lambda\left(\frac{q_j s_k}{T-a}\right).$$

There are rational functions $D_{jk}(a)\in E(a)$ with

$$H_{jk}(a)=r_{jk}(a)F_\lambda(a)+D_{jk}(a).$$

To see this, subtract $r_{jk\ell}(a)$ from $r_{jk\ell}(T)$ before dividing by $T-a$. For a fixed rational function $r$, the divided difference $(r(T)-r(a))/(T-a)$ is a finite sum of fixed rational functions of $T$ times rational functions of $a$. Apply $\lambda$ after multiplying by $b_\ell$. Thus each $D_{jk}$ depends linearly on finitely many fixed values of $\lambda$. Poles in the parameter $a$ are excluded below.

Use parameter pairs

$$\Omega=\{(a,b):a\in E\setminus\{0\},\ |a|\ne1, \ b=1/\bar a\}.$$

This set is Zariski dense in $E^2$. Indeed, the rational change of variables $a=s+it$, $b=1/(s-it)$ is invertible on a nonempty open set, and the grid $(s,t)\in F^2$, after removing finitely many polynomial zero sets, is Zariski dense over $E$. Put

$$U_{a,b}=\frac{T-b}{T-a}=1+\frac{a-b}{T-a}.$$

For $(a,b)\in\Omega$, the element $aU_{a,b}$ is a unit, since $|T|=1$. The cover applied to $s_kU_{a,b}$ gives, for each $k$, some $j$ such that

$$0=\lambda(q_j s_k)+(a-b) \bigl(r_{jk}(a)F_\lambda(a)+D_{jk}(a)\bigr).$$

For any assignment of these $j$'s, at least $d$ selected rows have the same $j$. Invert their matrix. After excluding finitely many additional values of $a$, this expresses $F_\lambda(a)$ in one of finitely many forms

$$F_\lambda(a)=R_\beta(a)+\frac{S_\beta(a)}{a-b},$$

where $R_\beta,S_\beta$ are rational vectors. Choose one valid branch at each parameter pair. The resulting finitely many branch sets cover the remaining part of $\Omega$.

We claim that every Zariski-dense branch has $S_\beta=0$, and that all Zariski-dense branches have the same $R_\beta$. For the first assertion, take two pairs $(a,b),(c,e)$ from the same dense branch and write $h=a-b$, $k'=c-e$. For a fixed $s_k$, apply the cover to $s_kU_{a,b}U_{c,e}$. The partial-fraction identity gives, for some $j$,

$$0=L_j+hH_j(a)+k'H_j(c) +\frac{hk'}{a-c}\bigl(H_j(a)-H_j(c)\bigr),$$

where $L_j=\lambda(q_j s_k)$ and, on this branch,

$$H_j(a)=V_j(a)+\frac{W_j(a)}h, \qquad V_j=r_{jk}R_\beta+D_{jk},\quad W_j=r_{jk}S_\beta.$$

The product of the displayed expressions over $j$ vanishes on a Zariski-dense subset of $E^4$. Therefore at least one factor is the zero rational function. In that factor compare coefficients in the independent variables $h,k'$. The coefficient of $hk'$ says $V_j(a)=V_j(c)$, so $V_j$ is constant. The coefficient of $h$ is $V_j(a)-W_j(c)/(a-c)$; varying $a$ forces both that constant and $W_j$ to vanish. The constant coefficient then gives $L_j=0$. For each $k$ we have therefore found a $j$ with $r_{jk}S_\beta=0$. Pigeonhole and the invertibility of any $d$ rows for one $j$ give $S_\beta=0$.

For two dense branches $\beta,\gamma$, repeat the argument using one pair from each. Both $S$'s vanish. A zero factor now has coefficients $L_j=0$, $r_{jk}R_\beta+D_{jk}=0$ and $r_{jk}R_\gamma+D_{jk}=0$. Again pigeonhole gives $R_\beta=R_\gamma$. This proves the claim.

There is at least one dense branch: finitely many proper algebraic subsets cannot cover a dense subset of $E^2$. Every nondense branch lies in a proper algebraic subset. Consequently there is a unique rational vector $R_\lambda(a)$ agreeing with $F_\lambda(a)$ outside a proper algebraic subset of the parameter plane. Moreover all possible $R_\lambda$, as $\lambda$ varies over functionals obeying the cover, lie in one fixed finite-dimensional subspace $W\subset E(a)^d$. For each branch the row inverse is fixed, and its other terms are linear in finitely many values of $\lambda$; take the sum of these finitely many spaces. In particular the polynomial degrees at infinity and pole orders of vectors in $W$ are bounded independently of $\lambda$. This construction used only the cover equation, so the same $W$ applies without the normalization $\lambda(1)=1$, including to the zero functional.

A finite-dimensional quotient

For every $v\in K_N$, the functional $\lambda_v(z)=\lambda(vz)$ obeys the same cover. Let $\Phi(v)=R_{\lambda_v}\in W$. Finite linear relations among units hold for their Cauchy transforms, and hence for their rational vectors: intersect the complements of the finitely many exceptional algebraic sets. This is still a dense parameter set. Thus $\Phi$ extends to an $E$-linear map $A\to W$.

Since $T$ is a unit, the exact Cauchy identity yields

$$\Phi(Tv)(a)=a\Phi(v)(a)+\Lambda(v), \qquad \Lambda(v)=\bigl(\lambda(vb_\ell)\bigr)_\ell \quad(v\in A).$$

Every $\Phi(v)$ is strictly proper at infinity. Otherwise its nonzero polynomial part, multiplied by $a^m$ in

$$\Phi(T^m v)=a^m\Phi(v)+ \sum_{j=0}^{m-1}a^{m-1-j}\Lambda(T^jv),$$

has degree at least $m$, while the sum has degree at most $m-1$. This contradicts the uniform bound on polynomial degrees in $W$ for large $m$. Comparing the constant polynomial parts in the one-step identity shows that $\Lambda(v)$ is minus the $a^{-1}$ coefficient of $\Phi(v)$. It is therefore determined linearly by $\Phi(v)$. Multiplication by $T$ consequently induces a linear operator on the finite-dimensional image of $\Phi$. A nonzero annihilating polynomial $p\in E[t]$ of that operator gives

$$\lambda(p(T)v)=0\qquad(v\in A),$$

using $b_1=1$. If the image is zero the same assertion holds with $p=1$.

The algebra $R=A/p(T)A$ is finite dimensional over $E$. For completeness, $\dim_E A/(T-a)A\le d$ for every $a\in E$. Any $d+1$ representatives in $A\subset N$ have a relation over $E(T)$. Clear denominators and divide all coefficients by their largest common power of $T-a$. Reduction modulo $(T-a)A$ gives a nontrivial relation over $E$. The filtration by powers of $T-a$ gives the corresponding bound for $(T-a)^m$, and factorization of $p$ over $E$, followed by the Chinese remainder theorem, gives $\dim_E R\le d\deg p$ when $p$ is nonconstant. The functional $\lambda|_A$ descends to an $E$-linear functional on $R$. If $R=0$, this already contradicts $\lambda(1)=1$; otherwise proceed as follows.

The cover is impossible in the quotient

Real elements of $A$ span $A$ over $E$, since $A$ is stable under conjugation. Choose real $z_1,\ldots,z_m\in A$ whose images span $R$. For $t=(t_1,\ldots,t_m)\in F^m$ put $z(t)=\sum t_i z_i$ and

$$u(t)=\frac{1+iz(t)}{1-iz(t)}\in K_N.$$

The denominator is invertible in $A$, since $(1-iz(t))^{-1}=(1+u(t))/2$. In the finite-dimensional algebra $R$, the coordinates of $u(t)$ are rational functions of $t$ with denominator nonzero on $F^m$. The cover therefore makes $\prod_j\lambda(q_j u(t))$ the zero rational function: $F^m$ is Zariski dense over $E$. One factor, say the $j$th, is identically zero. Its formal first derivatives at $t=0$ give

$$2i\lambda(q_j z_i)=0\qquad(1\le i\le m).$$

The images of $z_i$ span $R$, and $q_j$ is a unit of $R$. Hence the descended functional is zero on all of $R$. This contradicts $\lambda(1)=1$, and proves Theorem 1. $\square$