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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.

This asks whether the Jones polynomial equal to one characterizes the unknot among knots. A zero breadth of a Laurent polynomial makes $V_K$ a monomial; the identities $V_K(1)=1$ and $V_K^{\prime }(1)=0$ then force that monomial to be one. Thus using Jones breadth as a strictly positive complexity for nontrivial knots requires this detection property. To prove the derivative identity, differentiate the Jones polynomial skein relation at one for a crossing change in a knot. Both crossed diagrams have value one, while the smoothed two-component link has value minus two by Jones polynomial evaluation at one, so $V_{K_{+}}^{\prime }(1)=V_{K_{-}}^{\prime }(1)$. Crossing changes reduce to the unknot, whose derivative is zero. The corresponding Alexander polynomial detection property is false, as nontrivial untwisted Whitehead doubles demonstrate. The Jones question is discussed in research on closed four-strand braids.

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.

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.

Take the untwisted Whitehead double of a trefoil knot. The Whitehead pattern has winding number of a satellite pattern zero and becomes the unknot when its companion is the unknot. The Satellite formula for the Alexander polynomial therefore givesThe Whitehead pattern is geometrically essential in its solid torus, so the satellite knot with nontrivial companion $T_{2,3}$ is nontrivial. Hence $K_4$ is not isotopic to the unknot despite having trivial Alexander polynomial of a knot.

The unknot $U$ is the knot type represented by the planar circleEquivalently, an unknot is a knot that bounds a smoothly embedded disk in the three-sphere. The disk must itself be embedded; merely filling a loop with a map that is allowed to intersect itself does not establish that the knot is an unknot. Flattening a neighborhood of an embedded spanning disk explains why this criterion recovers the planar circle.

A complicated drawing can still represent an unknot. Starting with a crossing-free knot diagram of $U$, insert a small kink by a type I Reidemeister move. The resulting knot diagram has one crossing, but removing the kink exhibits the same unknot. Repeating in separate small regions gives knot diagrams of the unknot with arbitrarily many crossings, although the crossing number of a knot is $c(U)=0$. A drawing's apparent complexity is therefore not a certificate of knotting.

The normalized Alexander polynomial of a knot and Jones polynomial of the unknot are both $1$, and its knot group is the infinite cyclic group $\mathbb{Z}$, because the knot exterior is a solid torus. These are useful comparison values. In particular, a nonconstant Alexander polynomial of a knot certifies that a knot is not an unknot, but the converse fails: nontrivial untwisted Whitehead doubles also have Alexander polynomial of a knot $1$.