= Solution
Let F be an oriented genus-g <Seifert surface> for K, and choose a basis $a_1,\ldots,a_{2g}$ 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 is
$$
V_{ij}=\operatorname{lk}(a_i^+,a_j).
$$
The 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 $X=S^3\setminus\operatorname{int}N(K)$ for the <knot exterior>. Linking number with K gives the epimorphism $\pi_1(X)\to\mathbb Z$ sending a positive meridian to one. Let $\widetilde X$ be the associated <infinite cyclic cover>, with <deck transformation> t. The <Alexander module of a knot> is $\mathcal A_K=H_1(\widetilde X;\mathbb Z)$, 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 $\operatorname{lk}(b_j,a_i)=\delta_{ij}$. This <homology> statement does not require Y to be a handlebody. The two inclusions satisfy
$$
i_+(a_i)=\sum_jV_{ij}b_j,\qquad i_-(a_i)=\sum_jV_{ji}b_j.
$$
Indeed 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 $\operatorname{lk}(a_i^-,a_j)=\operatorname{lk}(a_j^+,a_i)$.
Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\widetilde X$. The <Mayer–Vietoris sequence>, with all translates collected into Laurent <modules>, contains
$$
H_1(F)\otimes\Lambda\xrightarrow{\,t i_+-i_-\,}H_1(Y)\otimes\Lambda\longrightarrow\mathcal A_K\longrightarrow H_0(F)\otimes\Lambda\xrightarrow{\,t-1\,}H_0(Y)\otimes\Lambda.
$$
The 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. Consequently
$$
\boxed{\Delta_K(t)\doteq\det(tV-V^T),\qquad\operatorname{br}\Delta_K\leq2g_s(K),}
$$
where $\doteq$ 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 is
$$
A_O\ B_U\ J_O\ I_U\ H_O\ F_U\ G_O\ H_U\ I_O\ J_U\ D_O\ C_U\ E_O\ D_U\ B_O\ A_U\ C_O\ E_U\ F_O\ G_U.
$$
The 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 are
$$
(o_0,\ldots,o_9)=(7,4,9,2,1,8,5,0,6,3).
$$
The <Wirtinger presentation> has $x_{k+1}=x_{o_k}x_kx_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently <Fox derivatives>, give the Alexander relation <matrix>
$$
M_{kj}=t\delta_{j,k}-\delta_{j,k+1}+(1-t)\delta_{j,o_k}.
$$
Delete 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 to
$$
N=\begin{pmatrix}th+(1-t)^2k&-h\\t(1-t)k&\ell\end{pmatrix}.
$$
For example, writing the relative generators as $z_0,\ldots,z_8$, the eliminated relations give $z_3=tz_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=hz_2$, $z_7=(1-t)hz_2+tz_6$, $z_0=t^{-1}(t^2-2t+2-(1-t)^2h)z_2-(1-t)z_6$, and $z_8=(1-t)kz_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. Hence
$$
\boxed{\Delta_K(t)\doteq\det N=5t^4-15t^3+21t^2-15t+5.}
$$
As 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 \b[K is prime].
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