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The Seifert genus is additive under connected sum of knots:The upper bound comes from boundary-connected-summing minimal Seifert surfaces, while the reverse inequality follows by cutting any Seifert surface for the connected sum along a splitting sphere.

A minimal-genus Seifert surface has a $2g_s(K)$-by-$2g_s(K)$ Seifert matrix $V$. Each entry of $tV-V^T$ has degree at most one, so its determinant has breadth of a Laurent polynomial at most $2g_s(K)$. This lower bound on Seifert genus need not be sharp: an untwisted Whitehead double can have Alexander polynomial of a knot $1$ and positive Seifert genus.

Boundary-connected-summing Seifert surfaces gives a block-diagonal Seifert matrix, because basis curves in the separated summands have zero linking. The Seifert-matrix presentation of the Alexander module is block diagonal too. Thus the Alexander modules form a direct sum and their Alexander polynomials of a knot multiply up to a unit.

A Hopf band is an annulus with one full twist whose boundary is a Hopf link. Plumbing it with another band adds a generator to the first homology of a Seifert surface. Its core has self-linking number $1$ or $-1$, depending on the twisting convention.

The linking number of two disjoint oriented closed curves in $S^3$ is the signed intersection number of one curve with a Seifert surface for the other. In an oriented diagram it is half the signed sum of crossings between the two components.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

The Seifert algorithm supplies a Seifert surface for every oriented knot: smooth the crossings in their oriented directions, span the resulting Seifert circles by disks at different heights, and restore each crossing with a half-twisted band. If needed, join components by tubes away from the boundary. Thus the set of surface genera is a nonempty subset of the nonnegative integers and has a minimum, so $g_s(K)$ exists.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum.

The same induction does not follow from the three suggested replacements. For crossing number of a knot, joining minimal diagrams gives only subadditivity; a reverse inequality giving strict descent for every nontrivial summand is not provided by this argument. The Jones polynomial of a connected sum does make its breadth of a Laurent polynomial additive, but the needed implication that zero breadth forces the unknot is the Jones unknot detection problem. For the Alexander polynomial, even this implication is false: nontrivial untwisted Whitehead doubles have Alexander polynomial one and hence breadth zero. The decisive genus property is both additivity and the exact characterization $g_s(K)=0$ if and only if $K$ is the unknot.

For uniqueness use complete decomposing sphere systems for a knot. Cut each system and cap the marked boundary spheres with straight arcs; the resulting oriented factors are prime knots. Compare two systems in transverse position, and use the prime-ball exchange lemma to simplify their intersection. Choose an innermost intersection circle bounding a disk in a sphere of one system and lying in a prime-factor ball of the other. If the disk misses the knotted arc it cuts off an empty ball. If it meets the arc once it cuts the prime factor into two knots, so one is an unknot. Replace the old ball by the side carrying the same prime factor, perturb its boundary slightly, and remove that intersection circle. This preserves the unordered factors: the discarded empty or trivial ball merely changes the joining arc, and any intervening capped boundaries still carry their original factors. If the chosen disk contains both marked points, take another innermost disk without marked points, or the complementary disk when the intersection circle is the only one.

Iteration yields an innermost prime-factor ball disjoint from both systems. Its arc has no nontrivial further decomposition, so it represents one factor in each. The boundary can be included in both systems using the same ball exchange; replace its arc by a straight boundary arc and remove this common factor. Induction on the number of factors matches the remaining factors. When the systems already have no intersections, start directly with an innermost ball and the same removal step. All replacements preserve the arc orientation, so the match is of oriented knots, not only unoriented types. Hence the prime decomposition of knots is unique up to reordering and oriented link isotopy.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

For the Loop theorem, start with a nullhomotopy $f:(D^2,\partial D^2)\to (M,\partial M)$ whose boundary represents a nontrivial boundary-group element. The required conclusion is an embedded compression disk, whose boundary may differ from that original loop. A tower proof proceeds as follows.

Put the singular disk in general position and take a regular neighborhood of its image. Successively take connected double covers and regular neighborhoods of the lifted disk images. Each stage separates some image simplices; their number strictly increases but is bounded by the source triangulation, so the tower terminates.

At the top there is no connected double cover, hence $H^1(-;\mathbb{Z}/2)=0$. Poincare-Lefschetz duality makes every boundary component a sphere. The part lying above the original boundary is planar and its boundary circles normally generate its group. Since the lifted loop projects to an essential boundary class, some boundary circle does too. Cap that circle on the sphere and push the disk inward.

Descend each double cover. The projected embedded disk has only double arcs and circles. Innermost disk exchanges and cut-and-paste surgeries remove these. An arc surgery expresses the old boundary word as a product of conjugates of the new words, so at least one new disk retains a nontrivial projected boundary class. Circle surgery retains that class. Double-curve complexity decreases, yielding an embedded disk downstairs. Iterating reaches the required compression disk in $M$.

Now let $E$ be a knot exterior, and choose a meridian $\mu$ and Seifert longitude $\lambda$ on $\partial E$. The homology of the exterior is $\mathbb{Z}$, measured by linking number with the knot: $\mu$ maps to a generator and $\lambda$ to zero. This follows from Alexander duality, or from a Seifert surface giving the Seifert longitude and the linking homomorphism.

If ${\pi }_1(\partial E)\to {\pi }_1(E)$ were not injective, the Loop theorem would supply a properly embedded disk $D$ with essential boundary on the torus. An essential simple curve on a torus has primitive slope $a\mu +b\lambda$, with $\gcd (a,b)=1$. Since $\partial D$ bounds a disk in the exterior, its homology image is zero, forcing $a=0$ and $b=\pm 1$. Thus $\partial D$ is a Seifert longitude. Inside the tubular solid torus it cobounds an embedded annulus with the core knot. Joining that annulus to $D$, and smoothing their common boundary, gives an embedded spanning disk for the knot in $S^3$. A knot bounding such a disk is the unknot. Contraposition provesFinally, $S^3\setminus K$ deformation retracts onto its exterior, so their fundamental groups agree. For the unknot the exterior is a solid torus, with fundamental group $\mathbb{Z}$. Conversely, if the knot-complement group were $\mathbb{Z}$, it could not contain an injected subgroup ${\pi }_1(\partial E)={\mathbb{Z}}^2$. The preceding result therefore forces the knot to be trivial. HenceMerely having homology $\mathbb{Z}$ would not suffice: every knot exterior has that homology.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum. Uniqueness is not needed here.

For coprime integers p,q, define the torus knot as the oriented parametrized curveCoprimality makes it an embedded single circle: two parameters with both coordinate phases equal differ by a multiple of $2\pi$. It lies on the standard unknotted torus separating the two solid tori. Negative exponents change the slope/orientation convention; they are allowed. If an exponent is zero, coprimality forces the other to be plus or minus one, giving an unknot.

Assume now $|p|,|q|\ge 2$. We give a group-theoretic proof that torus knots are prime, making the peripheral ingredient explicit. The Seifert-van Kampen theorem applied to the two solid-torus sides of the exterior, joined along the annulus parallel to the knot, givesHere a and b are the two core loops. The element $z=a^p=b^q$ is central. Under the abelianization $G\to \mathbb{Z}$, choose $a\mapsto q$, $b\mapsto p$; these generate the target because p,q are coprime. Thus $z\mapsto pq\ne 0$, so z is nontrivial. A meridian has, up to conjugacy, the form $m=a^u{b}^v$ with $qu+pv=1$. This follows by tracing the meridional disk across the two torus sides, using a dual primitive slope intersecting the knot slope once. Different Bezout solutions change u by a multiple of p and v by the opposite multiple of q, leaving the same word because z is central.

Quotienting by z gives $G/⟨z⟩=C_{|p|}\ast C_{|q|}$. Neither u nor v is zero modulo the corresponding cyclic order, by $qu+pv=1$. Hence the image of m is a cyclically reduced word of two syllables. All its nonzero powers remain reduced, so it has infinite order. In particular G is nonabelian and the knot is not an unknot.

We use the Loop theorem in its usual form: a noninjective boundary fundamental-group map of a three-manifold supplies an embedded compression disk. For a nontrivial knot, its boundary torus is incompressible. To see why, every sphere in its exterior bounds a ball on the side not containing the knot, by the Schoenflies theorem. A compression of the boundary torus would produce a sphere and make the exterior a solid torus. The compression disk has zero homology class in the exterior, so its boundary is the preferred longitude; attaching the annulus from that longitude to the knot gives a spanning disk and an unknot. This proves that nontrivial knot exteriors have incompressible boundary. Their peripheral groups inject as ${\mathbb{Z}}^2$, so their groups strictly contain the cyclic meridian subgroup.

If the torus knot were a connected sum of two nontrivial knots, van Kampen on the decomposing annulus would give $G=G_1{\ast }_{⟨m⟩}{G}_2$, with the meridian subgroup proper in each factor. The center of a proper amalgam lies in the amalgamated subgroup. One proof uses its Bass-Serre tree: a central elliptic element fixes a vertex orbit and its convex hull, which is the whole tree; a central translation would force the minimal tree to be its invariant axis, but a nontrivial vertex reflection reverses that translation. Thus a central element fixes every edge and lies in the edge subgroup.

The central element z would therefore equal $m^k$. Its image in $G/⟨z⟩$ would give ${\overline{m}}^k=1$, forcing $k=0$ because that image has infinite order. This contradicts $z\ne 1$. Consequently

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

Use ordinary integer surgery coefficients, so the positive trace has intersection form $(n)$. Let ${\hat{F}}_0\subset K_0$ be the capped Seifert surface. Label the Spin-c structures byOn the trace $X_n:S^3\to K_n$, let ${\hat{F}}_n^2=n$ and label its structures ${\mathfrak{r}}_j$ by $⟨c_1({\mathfrak{r}}_j),[{\hat{F}}_n]⟩=2j-n$. Restriction identifies ${\mathfrak{t}}_k$ with $j\equiv k(\mathrm{mod}n)$. In these compatible labels the Integer-surgery exact triangle in Heegaard Floer homology isHere boldface denotes the U-adic completion of Heegaard Floer homology. Only finitely many zero-surgery structures contribute, by the genus bound on a capped Seifert surface, so the middle direct sum causes no difficulty. The infinite sums of trace maps do require completion. The printed minus notation should be understood in this completed sense; with the polynomial version, a sum over every extension need not even have values in $\mathbb{Z}[U]$. The plus version has the same ordering and needs no completion.

The two ordinary surgery traces give $A_k$ and $C_k$. For $X_0:S^3\to K_0$, $H^2(X_0)=\mathbb{Z}$, and its Spin-c structures ${\mathfrak{q}}_i$ have determinant evaluations $2i$ on the capped Seifert surface. Their restrictions are the unique structure on $S^3$ and ${\mathfrak{s}}_i$ on $K_0$. Thus the component of $A_k$ in the $i$th summand is $F_{X_0,{\mathfrak{q}}_i}^{-}$ for $i\equiv k$. For the orientation-reversed and turned-around trace $V_n:K_n\to S^3$, $H^2(V_n)=\mathbb{Z}$ and ${\hat{F}}_n^2=-n$. Its structures ${\mathfrak{r}}_j$ still have characteristic evaluations $2j-n$, restrict to ${\mathfrak{t}}_{j\mathrm{mod}n}$ on $K_n$ and to the unique structure on $S^3$, andCharacteristic parity explains the allowed evaluations; changing a structure by the generator of $H^2$ changes its evaluation by two, and the boundary restriction reduces the label modulo $n$.

For $n>1$, $B_k$ has an auxiliary lens-space input. More precisely, take the three-boundary surgery cobordism, or equivalently a two-ended cobordism $Z_n:K_0\#L_n\to K_n$, where $L_n$ is positive $n$-surgery on the unknot. Its incoming lens-space factor is fixed to the distinguished structure ${\mathfrak{u}}_0$ and its canonical top Floer generator ${\Theta }_{L_n}$. ThenThis describes the boundary restrictions exactly, including the boundary that must not be suppressed. A concrete Kirby diagram for $Z_n$ starts with the split incoming surgeries of coefficients $0$ and $n$, then attaches a zero-framed circle linking each incoming component once. Its full linking matrix is $\left(\begin{array}{ccc}0& 0& 1\\ 0& n& 1\\ 1& 1& 0\end{array}\right)$. Quotienting cohomology by the two incoming surgery columns gives $H^2(Z_n)=\mathbb{Z}\oplus \mathbb{Z}/n$. Accordingly its Spin-c structures may be labeled $(i,q)$ with incoming restrictions $({\mathfrak{s}}_i,{\mathfrak{u}}_q)$ and outgoing restriction ${\mathfrak{t}}_{i-q}$. In particular, fixing $q=0$ gives exactly one extension for each $i$, and its outgoing class is $i\mathrm{mod}n$. Thus $B_k$ is a Heegaard Floer cobordism map with the fixed auxiliary generator, rather than an unqualified map of a trace from $K_0$ to $K_n$. When $n=1$, $L_n=S^3$ and all three arrows become the familiar sums of ordinary two-ended Heegaard Floer cobordism maps.