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Assume $g_s(K)>0$. A nontrivial connected sum of knots has positive genus summands, and additivity of Seifert genus applies. Its Alexander polynomial of a knot is the product of the summand polynomials. If the polynomial is irreducible, one summand has unit polynomial; the other must then have polynomial breadth $2g_s(K)$, exceeding twice its strictly smaller genus. This contradicts the Alexander breadth bound on Seifert genus. Irreducibility without the full-degree hypothesis alone does not imply primeness.

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.

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.

Number the seven crossings from top to bottom. Orient the strand entering the top-left arch downward. With the usual positive braid crossing convention, the crossing signs areTo make the diagram calculation reproducible, its oriented Gauss code isSubscripts record overpassing and underpassing. In this convention a positive trefoil knot has knot signature $-2$.

Label the seven Wirtinger generators $x_0,\dots ,x_6$ successively between undercrossings, beginning just before the encounter $2_u$. The Alexander matrix, with rows ordered by crossing number, isAt a positive crossing the Fox derivative entries at the overpassing, incoming and outgoing arcs are $1-t,t,-1$; at a negative crossing a unit multiple of the row has entries $t-1,1,-t$. Deleting the last row and column gives determinant $t(t^4-5t^3+7t^2-5t+1)$. Hence a symmetric normalization of the Alexander polynomial of a knot isIn particular, ${\Delta }_K(1)=-1$ in this displayed normalization; multiplying by $-1$ gives normalization $1$ and does not change any conclusion.

The Seifert algorithm produces four Seifert circles and seven bands. Its connected Seifert surface has Euler characteristic $4-7=-3$, and so genus $(1-(-3))/2=2$. Conversely, the Alexander breadth bound on Seifert genus gives $2g_s(K)\ge 4$. ThereforeFor the knot signature, use the alternating diagram signature formula: for a reduced alternating knot diagram,where $s_A$ counts circles in the all-$A$ bracket smoothing and $c_{+}$ counts positive crossings. Here the all-$A$ smoothing has four circles, the all-$B$ smoothing has five, and $c_{+}=5$. ThusThe opposite global knot signature convention gives $+2$ instead.

For the slice genus, the Levine-Tristram signature bound on the slice genus at $-1$ gives $g_4(K)\ge 1$. There is also an explicit unknotting crossing: switch crossing $3$. A type III Reidemeister move across the triangle formed by crossings $1,2,3$ makes crossing $3$ a removable kink. Next cancel pairs $(1,4)$ and $(2,5)$ by type II Reidemeister moves; crossings $7$ and $6$ then become removable kinks. This leaves the unknot. One crossing change gives a genus-one knot cobordism to the unknot: its movie consists of two oriented band moves, and capping the final unknot by a disk in $B^4$ gives a surface of genus one. Consequently

Choose an oriented Seifert surface $F$ of genus $g$ and an integral basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push $a_i$ in the positive normal direction to $F$. The Seifert matrix isIt represents the Seifert form. Its skew-symmetrization $V-V^T$ is the integral intersection form of $F$. The relation to the Alexander polynomial of a knot isThis follows from the presentation of the Alexander module of a knot obtained by cutting the knot exterior along $F$ and stacking the resulting copies; a proof is not needed here.

The printed genus inequality is false. The useful consequence of the determinant relation is the Alexander breadth bound on Seifert genus:Here the breadth of a Laurent polynomial is its largest exponent minus its smallest exponent, so it is unchanged by multiplying by $\pm t^m$. Indeed, for a minimal-genus Seifert surface, the matrix has size $2g_s(K)$ and each entry of $tV-V^T$ has degree at most one. Its nonzero determinant is an ordinary polynomial of degree at most $2g_s(K)$; its breadth is no larger than that degree. Equivalently, the highest exponent of a symmetrically normalized Alexander polynomial of a knot is at most $g_s(K)$. An unnormalized degree is not invariant under Laurent units.

For a counterexample to the printed inequality and the requested example, take the untwisted Whitehead double of a trefoil knot. The picture specifies the two-strand tangle in the zero Seifert framing; the two exterior caps form a Whitehead double clasp. The inset identifies the trefoil knot used as companion.

Its usual genus-one Seifert surface is a zero-framed annulus following the trefoil knot, joined by the clasp band. Choose the annulus core and a curve traversing the clasp band as the basis of its first homology. Zero annulus twisting gives the first self-linking number $0$. The clasp contributes a Hopf band with self-linking number $-1$ in the positive-clasp convention used here. Plumbing the bands contributes one linking in one push-off direction and none in the other. Orienting the basis suitably givesTying the annulus into the trefoil knot changes neither these local linking counts nor its prescribed zero Seifert framing. ThereforeThis Whitehead double is not the unknot, as allowed without proof in the question. Its Seifert genus is consequently at least one and at most one, hence exactly one. Thus $g_s(K)=1$ while every normalized constant Alexander polynomial of a knot has degree zero: a direct counterexample to the printed inequality.