Picard-one Fano index divisor is the integral generator

Let $n=\dim X$. The hypotheses $\rho (X)=1$ exclude $n=0$. Smooth connected projective $X$ is integral, so $H^0(X,O_X)=\mathbb{C}$. Empty-boundary adjoint vanishing from [F1] applied to $M=0$, with $M-K_X=-K_X$ ample, gives $H^i(X,O_X)=0$ for $i>0$. Consequently $\chi (O_X)=1$. This uses algebraic coherent cohomology and no analytic exponential sequence.

Let $N$ be any numerically trivial line bundle: its degree on every integral curve is zero. For each positive-dimensional integral subvariety $Y$ of dimension $k$, expand $(-K_X+N{)}^k\cdot Y$. Every term containing $c_1(N)$ vanishes, since after removing one such factor the remaining intersection is a one-cycle, a finite rational linear combination of integral curves. Thus $(-K_X+N{)}^k\cdot Y=(-K_X{)}^k\cdot Y>0$. The numerical criterion [F2] makes $-K_X+N$ ample. Empty-boundary adjoint vanishing therefore gives $H^i(X,N)=0$ for all $i>0$. GRR to a point and $\mathrm{ch}(N)=\exp (c_1(N))$ from [F1] give $\chi (N)=\chi (O_X)=1$: every positive $c_1(N)$ term in the top-degree product $\mathrm{ch}(N)\mathrm{td}(T_X)$ again pairs $c_1(N)$ with a rational one-cycle and vanishes. Hence $h^0(X,N)=1$. A nonzero section has either no zero divisor, trivializing $N$, or an effective nonzero divisor $D$. In the latter case, for any ample Cartier $B$, $D\cdot B^{n-1}>0$ (take a very ample power and count the positive degrees of the components), contradicting numerical triviality. Thus every numerically trivial line bundle is trivial. In particular there is no torsion.

Fix an ample Cartier divisor $B$ and consider the group homomorphism $\delta :\mathrm{Pic}(X)\to \mathbb{Z}$, $N\mapsto c_1(N)\cdot B^{n-1}$. The numerical divisor group injects into the product of its integer curve-degree groups; it is therefore torsion free and injects into its real scalar extension. By $\rho (X)=1$, that real numerical space is one-dimensional. The functional $\delta$ on it is nonzero because $\delta (B)=B^n>0$, hence injective. If $\delta (N)=0$, $N$ is numerically trivial and is trivial by step 2.1. Thus $\delta$ is injective on $\mathrm{Pic}(X)$. Its image is a nonzero subgroup of $\mathbb{Z}$, so is $d\mathbb{Z}$ for its least positive element $d$. Choose $C$ with $\delta (C)=d$; injectivity proves $\mathrm{Pic}(X)=\mathbb{Z}C$. No finite-generation, Morse-cell, GAGA or Neron--Severi finiteness theorem is used.

The numerical class of $C$ is the positive rational multiple $\lambda =\delta (C)\mathrm{/}\delta (B)$ of the class of $B$. Mixed intersection numbers depend only on numerical divisor classes, by the same one-cycle expansion used in step 2.1. Thus $C^k\cdot Y={\lambda }^k{B}^k\cdot Y>0$ for every positive-dimensional integral $Y$, so [F2] makes $C$ ample. Write $A=kC$ in the integral Picard group. Since $A$ and $C$ are ample, $k>0$. If $k>1$, then $-K_X=(rk)C$ contradicts maximality of the stated index $r$. Hence $k=1$, and $\mathrm{Pic}(X)=\mathbb{Z}A$ integrally.

This proves exactly the primitive-generator input needed for the rank-one index-two Fano fourfold and rank-one index-one Fano threefold branches. No rational connectedness, simple connectivity, Fano classification or Picard-torsion theorem is imported.

The same rational-Chow argument gives any fixed torsion line Chern character equal to one: if $N^m=O$, then mc1(N)=0 and ch(N)=exp(c1(N))=1 after tensoring with Q. No numerical versus rational-equivalence substitution is made.