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Repeated Smith normal form puts a chain complex of finitely generated free abelian groups into its elementary decomposition of a finite free chain complex: one-term summands $\mathbb{Z}$ and two-term summands $0\to \mathbb{Z}\stackrel{m}{\to }\mathbb{Z}\to 0$. Applying ${\mathrm{Hom}}_{\mathbb{Z}}(-,\mathbb{Z})$ reverses a two-term summand but keeps the same multiplication by $m$. Reading its homology gives the universal coefficient theorem for cohomologysplit noncanonically. Thus if $H_n(C)\cong {\mathbb{Z}}^{b_n}\oplus T_n$ with $T_n$ finite, then

The universal coefficient theorem for homology with ${\mathbb{F}}_p$ coefficients givesIf the middle group vanishes for every prime $p$, so does the tensor term. Any nonzero free summand survives for every $p$, and any nonzero finite summand survives for a prime dividing its order. Since integral homology is finitely generated, it must vanish in every degree. This is detection of integral acyclicity modulo primes.

For the displayed complex, writeUsing $f_{\#}{d}_C=d_D{f}_{\#}$, one obtainsThus $M$ is the mapping cone of $f_{\#}$ in this sign convention. Its long exact sequence in homology shows that $f_{\ast }$ is an isomorphism exactly when $H_{\ast }(M)=0$. If $f_{\ast }$ is an isomorphism with ${\mathbb{F}}_p$ coefficients for every prime, then $M\otimes {\mathbb{F}}_p$ is acyclic for every $p$. Prime-field detection makes $M$ integrally acyclic, so the integral long exact sequence gives

The Smith normal form of an integer $m\times n$ matrix $A$ is a diagonal matrixfor whichwith $U\in G{L}_m(\mathbb{Z})$ and $V\in G{L}_n(\mathbb{Z})$. It exists and is unique.

The structure theorem for finitely generated modules over a principal ideal domain, specialized to $\mathbb{Z}$, says that every finitely generated abelian group has a unique invariant-factor decompositionEquivalently, its finite part is a direct sum of cyclic groups of prime-power order.

At the generic point of $L_0$, the functions $x_1,x_2$ are units and the equation gives $x_0=x_3^3/(x_1{x}_2)$. Taking $x_3$ as a uniformizer yieldsAt $L_1$, the functions $x_0,x_2$ are units and $x_1=x_3^3/(x_0{x}_2)$, so ${\nu }_{L_1}(x_0/x_3)=-1$. Symmetrically, ${\nu }_{L_2}(x_0/x_3)=-1$.

The complement of the three lines is $U=D_{+}(x_3)$, withThis unique factorization domain has trivial divisor class group, so part b says that $[L_0],[L_1],[L_2]$ generate $\mathrm{Cl}(X)$. The units on $U$ are scalar multiples of $u_0^a{u}_1^b$, and their boundary valuation vectors are generated byThe resulting integer matrix has Smith normal form $\mathrm{diag}(1,3,0)$. Consequently

$A={\mathbb{Z}}^2/⟨(6,9)⟩$. The class of $(2,3)$ has order $3$, so $A$ has torsion and is not free. A unimodular change of basis (Smith normal form, using $\gcd (6,9)=3$) gives $A\cong \mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$.

Map $X_{d,f}$ to $K(\mathbb{Z},2)={\mathbb{CP}}^{\infty }$ using its degree-two generator. The homotopy fibre is the total space of the corresponding circle bundle, is 2-connected, and has ${\pi }_3$ equal to that of $X_{d,f}$. Over $S^2\vee S^3$ its total space is $P=S^3{\cup }_{S^1}(S^1\times S^3)$, with $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$. The four-cell bundle is trivial; its relative degree-four generator has boundary $dA+fB$, the lifted attaching map. The long exact sequence in relative homology and Hurewicz theorem therefore give ${\pi }_3(X_{d,f})={\mathbb{Z}}^2/⟨(d,f)⟩$. Smith normal form gives the displayed expression when $d>0$, including $f=0$.