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For an integer matrix $M$, its $j$th determinantal divisor is the nonnegative greatest common divisor of all $j$-by-$j$ matrix minors, with ${\Delta }_0=1$. If all these minors vanish, take ${\Delta }_j=0$. Unimodular row and column operations preserve these divisors. For nonzero invariant factors $d_1\mid \cdots \mid d_r$ in the Smith normal form, ${\Delta }_j=d_1\cdots d_j$ for $j\le r$, so $d_j={\Delta }_j/{\Delta }_{j-1}$.

Over the principal ideal domain $\mathbb{Z}$, Smith normal form decomposes a chain complex of finitely generated free modules into direct sums of one-term complexes $\mathbb{Z}$ and two-term complexesThe one-term summands record free homology, the summands with $m>1$ record cyclic torsion, and those with $m=1$ are acyclic.

Glue the boundary of a Möbius band to the projective line in the real projective plane by a mapping degree $n$. The band is the mapping cylinder of the degree-two boundary covering of its core. Collapsing the connecting tree in the resulting double mapping cylinder gives a CW complex with one vertex, edges $a,b$, and two attaching words $a^2$ and $a^{-n}{b}^2$. Its cellular chain complex has degree-two boundary matrix $\left(\begin{array}{cc}2& -n\\ 0& 2\end{array}\right)$. Its determinant is four, so $H_2=0$. The Smith normal form gives $H_1\cong \mathbb{Z}/4$ for odd $n$ and $(\mathbb{Z}/2{)}^2$ for even $n$; $H_0=\mathbb{Z}$ and higher homology vanishes. The space cannot be homotopy equivalent to a closed manifold: mod-two homology forces a potential manifold to be a surface, and its Euler characteristic is one, whereas the only such closed connected surface is the projective plane, with different integral $H_1$.

For a square integer matrix $A$ with nonzero determinant, the image $A{\mathbb{Z}}^n$ is a subgroup of the integer lattice. Its index of a subgroup isIn particular, $A=\mathrm{diag}(2,1)$ gives all points with even first coordinate. The quotient has two cosets, represented by $(0,0)$ and $(1,0)$, and the fundamental cell has area two. A determinant of two therefore counts how much integer information is missing from the image.

The integrality hypothesis is essential. The rational matrix $R=\mathrm{diag}(2,1/2)$ has determinant one but sends $(0,1)$ to $(0,1/2)$. Its generated Euclidean lattice has the same covolume as ${\mathbb{Z}}^2$ without being the same integer lattice. The unimodular matrix criterion combines an arithmetic condition with a geometric one.

A short proof of the index formula uses Smith normal form. Write $PAQ=\mathrm{diag}(d_1,\dots ,d_n)$, with $P,Q$ unimodular matrices and positive integers $d_i$. They preserve the relevant integer structure, so the quotient has the same size as ${\prod }_i\mathbb{Z}/d_i\mathbb{Z}$, namely ${\prod }_i{d}_i=|\det A|$. No assumption that the matrix is orthogonal or its basis vectors are short is needed.

For a two-by-two matrix over a principal ideal domain, its first Smith normal form invariant is the greatest common divisor of its entries, while the product of its two invariants is its determinant, up to a unit. Hereso the entries generate the unit ideal in the Gaussian integers. The first invariant is therefore $1$. AlsoHence a valid Smith normal form isMultiplying the second diagonal entry by a unit $\pm 1,\pm i$ gives an equivalent normal form. The divisibility condition holds because $1$ divides every Gaussian integer.

Let $R={\mathcal{O}}_k$, $S={\mathcal{O}}_K$ and $n=[K:k]$. For each prime ideal $\mathfrak{P}$ of $S$ above $\mathfrak{p}$ of $R$, set $f_{\mathfrak{P}}=[S/\mathfrak{P}:R/\mathfrak{p}]$. The relative norm of a fractional ideal can be specified on the unique prime ideal factorization byHere is why this definition is the intrinsic relative ideal norm from norms of elements, and why the residue weights occur. Localize at $\mathfrak{p}$. The finite torsion-free $R_{\mathfrak{p}}$-module $S_{\mathfrak{p}}$ is free of rank $n$, because $R_{\mathfrak{p}}$ is a discrete valuation ring. Multiplication by a nonzero integral $a$ has determinant $N_{K/k}(a)$. The Smith normal form givesDecompose the quotient at the prime ideals above $\mathfrak{p}$. In $S_{\mathfrak{P}}$ each quotient ${\mathfrak{P}}^j/{\mathfrak{P}}^{j+1}$ is one-dimensional over $S/\mathfrak{P}$, so has length $f_{\mathfrak{P}}$ over $R_{\mathfrak{p}}$. ThereforeFor an integral ideal $I$, all norms of elements of $I$ have valuations at least $\sum f_{\mathfrak{P}}{v}_{\mathfrak{P}}(I)$. These minima can be attained simultaneously at the finitely many prime ideals above a fixed $\mathfrak{p}$. In fact the Chinese remainder theorem for ideals, applied inside the finitely generated module $I$, chooses an $a\in I$ with nonzero image in each $I/\mathfrak{P}I$. Equivalently $v_{\mathfrak{P}}(a)=v_{\mathfrak{P}}(I)$ at all these primes. The generated ideal of the element norms consequently has exactly the displayed valuations. Clearing a denominator extends this description to fractional ideals.

Multiplication of fractional ideals adds their prime valuations. The displayed formula therefore provesIt also gives $N_{K/k}((a))=(N_{K/k}(a))$.

Finally, reduction of the free rank-$n$ module $S_{\mathfrak{p}}$ modulo $\mathfrak{p}$ has dimension $n$ over $R/\mathfrak{p}$. The factorization $\mathfrak{p}S={\prod }_i{\mathfrak{P}}_i^{e_i}$ and the Chinese remainder theorem for ideals identify this reduction with the product of the quotients by ${\mathfrak{P}}_i^{e_i}$. Their filtrations have $e_i$ successive quotients of dimension $f_i$ over $R/\mathfrak{p}$. Adding dimensions proves the fundamental identity for prime decomposition:

In the integer basis $u,v,e_3$, direct substitution gives $w=-7u+5v+18e_3$. ThereforeThe first two basis coordinates disappear in the quotient and the third is taken modulo eighteen. Equivalently the Smith normal form has diagonal entries $1,1,18$.

Use additive notation for this finitely generated abelian group. The change of generators $A=a+3d$, $e=c-b$, together with $b,d$, is invertible over the integers: $a=A-3d$ and $c=b+e$. The second relation gives $2e=0$. Subtracting it from the first gives $4A=0$, while the third becomes $4b+2e=0$, hence $4b=0$. The last relation becomes $8A+4b+4e=0$ and is consequently redundant. Conversely $4A=4b=2e=0$ imply every original relation, so no further relation has been introduced or lost. Thus the cyclic group decomposition isThe infinite cyclic group factor contributes only zero to an element killed by two. Each finite factor has two such elements, so the kernel of multiplication by two has $2^3=8$ elements. Removing the identity leavesThe invertible generator change also supplies a direct presentation proof of the decomposition, without relying solely on a Smith normal form computation.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

There are no one-handles, so attaching the three two-handles to a four-ball cannot create a fundamental group generator: ${\pi }_1(W)=1$. The handle chain complex has $C_2={\mathbb{Z}}^3$, $C_1=0$ and $C_0=\mathbb{Z}$, givingTo calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving linking number $-3$. Each long component has two positive crossings with the small component, giving linking number $1$. The surgery linking matrix, in the order of framings $4,3,2$, is consequentlyChanging a component orientation changes the corresponding row and column signs, and does not change the resulting homology. The long exact sequence of a pair and Poincare-Lefschetz duality identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\mathrm{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The Smith normal form therefore has diagonal $1,1,7$, andIn particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.

For a basis of a module $m_1,\dots ,m_d$ of the full-rank module $M$, define its discriminant of elements of a number field byThe field discriminant is this determinant for an integral basis of ${\mathcal{O}}_K$. A change of integral basis with determinant $\pm 1$ does not change it. Write $m_j={\sum }_i{b}_{ij}{e}_i$ in an integral basis and let $B=(b_{ij})$. The two trace matrices satisfy $T_M=B^{\mathsf{T}}{T}_{{\mathcal{O}}_K}B$. The index of the lattice is $r=|\det B|$, as follows for example from Smith normal form. ThereforeMultiplicative closure makes $M$ an order when it contains one, but the determinant argument itself requires only full finite index, so it also covers the stated module hypothesis without that extra condition.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence

Let $E=\mathbb{C}/\Lambda$, and let ${\Lambda }^{\prime }$ be the inverse image of the cyclic subgroup $G$ under the quotient map. Then $\Lambda \subset {\Lambda }^{\prime }$ is an inclusion of Euclidean lattices with ${\Lambda }^{\prime }/\Lambda \cong \mathbb{Z}/N\mathbb{Z}$. The Smith normal form supplies a basis $u,v$ of ${\Lambda }^{\prime }$ for whichChoose the orientation of this basis so that $\mathrm{Im}(v/(Nu))>0$. Dividing the coordinate on $\mathbb{C}$ by $Nu$ gives an isomorphism of complex toriHere $u$ maps to $1/N$, and its class generates ${\Lambda }^{\prime }/\Lambda$. Thus the isomorphism carries $G$ to the required cyclic subgroup. Every pair is represented: $\Phi$ is surjective. This argument avoids assuming that an arbitrary integer lift of a primitive vector modulo $N$ is itself primitive.

The group presentation gives the quotient group ${\mathbb{Z}}^3$ by the subgroup generated by the rows ofFor the Smith normal form, the first determinantal divisor is the greatest common divisor of all entries, namely $3$. The nonzero two-by-two matrix minors are $\pm 27$, so the second determinantal divisor is $27$. The third row is the second minus the first, and a nonzero minor shows the rank is exactly two. Consequently the nonzero invariant factors are $3$ and $27/3=9$, with one free generator. The Fundamental theorem of finitely generated abelian groups givesAn element killed by $3$ has zero free coordinate, arbitrary coordinate in $\mathbb{Z}/3\mathbb{Z}$, and coordinate $0,3$ or $6$ in $\mathbb{Z}/9\mathbb{Z}$. There are nine such elements. Removing the identity leaves eight elements of order three.

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.

Choose the Hopf map $\eta$ and orientations so its Hopf invariant is one. The cellular cohomology of $Y_d$ is $\mathbb{Z}$ in degrees $0,2,4$ and zero otherwise. Let $x,z$ be generators in degrees two and four. Precomposing $\eta$ with a degree-$d$ self-map of $S^3$ represents $d\eta$. The induced map of mapping cones is the identity on the two-cell and has degree $d$ on the four-cell. Since the Hopf mapping cone has cup square equal to its top generator, naturality of the cup product gives $x^2=dz$. Equivalently, precomposition scales the Hopf invariant by degree. HenceThe coefficient is $d$, not $d^2$: multiplication in ${\pi }_3(S^2)$ is being used, rather than postcomposition by a degree-$d$ map of the target sphere.

For $X=X_{d,f}$, the only nonzero cellular boundary is $C_4\to C_3$, multiplication by $f$. Thus its cellular cohomology givesLet $x$ again generate $H^2$, and let $z$ be the class of the four-cell cochain. Collapsing the three-sphere gives a map $X\to Y_d$ which pulls the degree-two generator back to $x$ and the degree-four generator back to $z$. Hence $x^2=dz$, now interpreted modulo $f$. Every other product of positive-degree classes vanishes by dimension. This is the cohomology ring of a Hopf attachment with a sphere summand. More explicitly, if $f\ne 0$,If $f=0$, there is additionally a generator $y$ of degree three, with $xy=y^2=yz=0$. The displayed description of the groups uses $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so the zero case is included. In particular, a nonzero $f$ kills the degree-three cohomology but can leave torsion in degree four.

There are no one-cells, and attaching cells of dimension at least three does not change ${\pi }_1$. Thus $X$ is simply connected. The Hurewicz theorem in degree two gives

To compute ${\pi }_3$, choose a map $g:X\to {\mathbb{CP}}^{\infty }=K(\mathbb{Z},2)$ representing $x$. This Eilenberg–MacLane space can be realized by the classifying space of $S^1$. On $S^2$ the map is the standard inclusion; on $S^3$ it is constant, and it extends over the four-cell since ${\pi }_3(K(\mathbb{Z},2))=0$. Its homotopy fibre $T$ is the pullback of the universal circle bundle $S^{\infty }\to {\mathbb{CP}}^{\infty }$. The map on ${\pi }_2$ is an isomorphism, so the long exact sequence of homotopy groups of a fibration gives ${\pi }_1(T)={\pi }_2(T)=0$ and ${\pi }_3(T)\cong {\pi }_3(X)$. Therefore the Hurewicz theorem identifies the latter with $H_3(T;\mathbb{Z})$.

Over $W=S^2\vee S^3$, the circle-bundle total space isThe first summand is the Hopf fibration total space over $S^2$, and the second is the trivial bundle over $S^3$. Their intersection is the fibre over the wedge point. The Mayer–Vietoris sequence gives $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$, with generators from the two three-spheres. These are the lifts of ${\eta }_1$ and $i_2$. The circle bundle over $W$ also has a 2-connected total space, so lifting and the Hurewicz theorem identify $d{\eta }_1+f{i}_2$ with $dA+fB$.

Over the attached four-disc the bundle is trivial. The resulting relative pair has the homology of $(S^1\times D^4,S^1\times S^3)$, so $H_4(T,P)=\mathbb{Z}$ and $H_3(T,P)=0$. Its boundary map sends a generator to the lift of the attaching map, namely $dA+fB$. The long exact sequence in relative homology now givesApplying Smith normal form to this one relation proves the third homotopy group of a Hopf attachment with a sphere summand:Since $d>0$, the greatest common divisor is positive even when $f=0$. This derivation uses the required map to an Eilenberg–MacLane space and determines the extension, rather than merely the orders of its pieces.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

A Smith normal form of an $m\times n$ matrix $A$ over a principal ideal domain $R$ is a diagonal matrixwhere $U\in G{L}_m(R)$ and $V\in G{L}_n(R)$. It is guaranteed to exist for every matrix over a principal ideal domain. The entries $d_i$ are determined up to multiplication by units; over $\mathbb{Z}$ they are conventionally chosen positive.

Present a finitely generated abelian group $G$ as the cokernel of an integer matrix $A:{\mathbb{Z}}^m\to {\mathbb{Z}}^n$. Changing $A$ by the invertible row and column operations used in its Smith normal form does not change the isomorphism type of the cokernel. Ifthenwhere factors with $d_i=1$ may be omitted. This is the Fundamental theorem of finitely generated abelian groups; the rank of an abelian group $r$ and invariant factors $d_i>1$ are unique.