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A nonzero integer polynomial of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of the Fredholm series. Functional transcendence makes the resulting analytic function nonzero. If $\alpha$ and $f_l(\alpha )$ were both algebraic, the relation $f_l({\alpha }^{l^r})=f_l(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}$ would keep the auxiliary values in one fixed number field. Their Absolute logarithmic Weil height is at most $N{l}^{r}h(\alpha )(1+1/(l-1))+O_N(r+1)$, whereas their chosen absolute value is $O_N(|\alpha {|}^{T_N{l}^r})$. The Liouville height inequality forces $T_N=O(N)$, contradicting its quadratic lower bound for sufficiently large $N$.

We prove the Mahler alternative in full. Let $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. Its series converges locally uniformly in the unit disc and satisfiesIt is transcendental as a function. At a root of unity $\zeta$ with ${\zeta }^{l^r}=1$, the radial tail ${\sum }_{n\ge r}{t}^{l^n}$ in $f(t\zeta )$ is real and tends to infinity as $t\uparrow 1$, while the initial terms remain bounded. These roots are dense on the unit circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and polynomial discriminant: elsewhere its defining polynomial and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Assume now that $\alpha$ and $f(\alpha )$ are algebraic, with $0<|\alpha |<1$, and put both in a number field $K$ of degree $d$. For each $N$, the $(N+1{)}^2$ coefficients of a polynomial $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have integer coefficients, so a nonzero integer solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge (N+1{)}^2-1$.

At ${\alpha }_r={\alpha }^{l^r}$ the functional equation gives $f({\alpha }_r)=f(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}\in K$. Thus ${\eta }_r=R_N({\alpha }_r)\in K$ is nonzero for all sufficiently large $r$ andUse the absolute logarithmic height $h$. The elementary inequalities for sums and polynomial evaluation giveHere $h(f({\alpha }_r))\le h(f(\alpha ))+(l^r-1)h(\alpha )/(l-1)+r\log 2$. The Liouville height inequality gives $\log |{\eta }_r|\ge -dh({\eta }_r)$. Divide by $l^r$ and let $r\to \infty$ to obtainThe left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence every such algebraic $\alpha$ has transcendental $f(\alpha )$. This is the auxiliary-value height argument for the Fredholm series.

For the rational continuation, let $a_n={\sum }_{j=0}^n(p/q{)}^{l^j}$ and $Q_n=q^{l^n}$. This is a rational number with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,A rational value $a/b$ would instead have distance at least $1/(b{Q}_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $ϵ>0$ with $2+ϵ<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-ϵ}$. This contradicts Roth theorem. Thus the requested rational values are transcendental by the Roth criterion for rational Fredholm values.

For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/{\mathbb{F}}_q$, then $\mathrm{deg}\Phi =q$ and $\mathrm{deg}(1-\Phi )=N=\#E({\mathbb{F}}_q)$: the latter map is separable and its kernel is exactly the rational points. The divisor identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\mathrm{deg}(u+v)+\mathrm{deg}(u-v)=2\mathrm{deg}u+2\mathrm{deg}v$. Polarizing it yieldsAll these degrees are nonnegative. In particular $\mathrm{deg}(-t+2\Phi )=4q-t^2\ge 0$, so $|t|\le 2\sqrt{q}$. The Frobenius characteristic polynomial is $X^2-tX+q$; its complex roots have modulus $\sqrt{q}$, including the double-root case. The zeta function has numerator $1-tT+q{T}^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the degree-form proof of the Hasse bound; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.

If $\alpha \ne 0$ has degree $d$, then at every Archimedean embeddingIndeed, the factor of the defining product belonging to that Archimedean place shows $\max (1,|\alpha |{)}^{1/d}\le H(\alpha )$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to ${\alpha }^{-1}$ and use $H({\alpha }^{-1})=H(\alpha )$ to obtain the lower bound. This is the Liouville height inequality.

Take distinct $x,y\in A_n$. Their difference has the formwhere $P\in \mathbb{Z}[X]$ has degree at most $n-1$ and polynomial length at most $nh$. By the height bound for a polynomial evaluation,The algebraic number $x-y$ is nonzero and has degree at most $d$, so the Liouville height inequality gives the separation

All elements of $A_n$ lie in an interval of length at mostSince $H(1-\alpha )\le 2H(\alpha )$, another application of the Liouville height inequality givesThe number of points in an interval is at most one plus its length divided by their minimum separation. ConsequentlyThus the requested statement holds, for example, with the absolute constant $C=2$.

For $\beta \in K$, the nonzero algebraic number $\gamma =\alpha -\beta$ has degree at mostThe basic height inequality givesThe Liouville height inequality therefore yieldsThus one may take the explicit constant