Finite-order smooth line bundles have torsion reduced bundle class

Choose finitely many line-bundle trivializing charts with smaller precompact subcharts covering $X$. In each larger chart choose a smooth nonnegative Euclidean bump ${\rho }_i$, supported inside it and positive on the smaller closure. Their sum is positive, so divide by the sum to make ${\sum }_i{\rho }_i=1$. A local nonzero frame $v_i$ defines a global smooth section $s_i={\rho }_i{v}_i$ by zero extension; the support is strictly inside its frame chart. At every point some $s_i$ is nonzero. Thus $s=(s_1,\dots ,s_m)$ is a nowherezero section of $L^{\oplus m}$. A Hermitian metric on $L$ is constructed by the same finite positive weighted sum of local squared frame norms, which is positive on every nonzero fiber vector. Normalize $s$ in the induced direct-sum metric. Its span is a trivial smooth line subbundle, and the smooth orthogonal projection $u\mapsto ⟨u,s⟩s$ exhibits a smooth complementary bundle $Q$ (in a local frame, extend the nonzero vector to a frame using a nonzero coordinate minor). Hence $L^{\oplus m}\cong \mathbf{1}\oplus Q$. For empty $X$ take $m=1$ and the zero bundle ring.

For actual bundles define ${\lambda }_{-1}(E)={\sum }_{k=0}^{\mathrm{rank}E}(-1{)}^k[{\bigwedge }^{k}E]$, componentwise if necessary. The fiberwise map which wedges an $i$-fold wedge from $E$ before a $j$-fold wedge from $F$ gives the smooth isomorphism ${\bigwedge }^k(E\oplus F)\cong {⨁}_{i+j=k}{\bigwedge }^{i}E\otimes {\bigwedge }^{j}F$. Local basis wedges prove bijectivity and respect transitions, so finite distributivity gives ${\lambda }_{-1}(E\oplus F)={\lambda }_{-1}(E){\lambda }_{-1}(F)$. For a line ${\lambda }_{-1}(L)=1-[L]$ and ${\lambda }_{-1}(\mathbf{1})=0$. Apply this actual-bundle identity to step1.1 to obtain $(1-[L]{)}^m={\lambda }_{-1}(L^{\oplus m})={\lambda }_{-1}(\mathbf{1}\oplus Q)=0$. No exterior-power operation on virtual differences or splitting principle is used.

Put $x=[L]-1$ in the commutative ring of [F1]. The given tensor-power isomorphism implies $(1+x{)}^d=1$. Expanding the finite binomial gives $dx=-x^{2}q(x)$, where $q(x)={\sum }_{j=2}^d\left(\genfrac{}{}{0px}{}{d}{j}\right)x^{j-2}$ (zero if $d=1$). Induction then gives $d^{k}x=(-1{)}^k{x}^{k+1}q(x{)}^k$ for every nonnegative integer $k$: multiply the preceding identity by $d$ and substitute $dx=-x^{2}q(x)$, using commutativity. Step2.1 says $x^m=0$, so taking $k=m-1$ gives $d^{m-1}x=0$, including $m=1$. This is an actual integer annihilator, not just a rational equality.

Applying an additive $I$ to step3.1 gives $d^{m-1}(I([L])-I(1))=0$ in $\mathbb{Z}$. The coefficient is a positive integer, so the difference is zero. All charts, sections, bundle splittings, powers and sums used above were finite; no triangulation, classification of line bundles, Chern character or infinite-dimensional operator argument entered the assertion.