The Alexander polynomial does not determine the prime knot summands

By [F1] and [F2], ${\mathrm{\Delta }}_T(t)\doteq \det (V-t{V}^T)=\det \left(\begin{array}{cc}t-1& t\\ -1& t-1\end{array}\right)=(t-1{)}^2+t=t^2-t+1.$ For $-T^{\ast }$ use the reflected spanning surface with its surface orientation reversed, so its boundary is the reversed reflected knot. Its positive normal is the reflected original positive normal: ambient reflection and surface reversal each negate the orientation, so their effects cancel in the normal-first convention. Reflection negates linking. The transported homology basis on this actual surface consequently has matrix $-V$. Its determinant is $\det (-V+t{V}^T)=\det (V-t{V}^T)$, since the matrix size is two. The two determinants evaluate to $3$ at $t=-1$, so this parameter is nonroot and [F3] applies directly, with no root-average ambiguity.

By [F2], the actual boundary connected-sum surfaces of $G=T\mathrm{\#}T$ and $Q=T\mathrm{\#}(-T^{\ast })$ have matrices $V\oplus V$ and $V\oplus (-V)$. Their block determinants both equal $(t^2-t+1{)}^2$, proving the two Alexander-polynomial identities up to units. Also $H_{-1}(V)=2(V+V^T)=\left(\begin{array}{cc}-4& -2\\ -2& -4\end{array}\right)\mathrm{.}$ Its eigenvalues on the vectors $(1,1)$ and $(1,-1)$ are $-6$ and $-2$, respectively. Thus ${\sigma }_{-1}(T)=-2$ and ${\sigma }_{-1}(-T^{\ast })=2$. The corresponding block forms give ${\sigma }_{-1}(G)=-4$ and ${\sigma }_{-1}(Q)=-2+2=0$. These are actual knot-invariant values by [F3], since the sum polynomials at $-1$ equal $9\mathrm{\ne }0$. In particular $\mathrm{\mid }{\sigma }_{-1}(G)\mathrm{\mid }=4$ as stated.

By [F4], $T$ and $T^{\ast }$ are nontrivial prime oriented knots. Reversal of knot orientation also preserves these properties: reverse both factors and the two local oriented arcs in any purported nontrivial connected-sum presentation of $-T^{\ast }$ to obtain a nontrivial presentation of $T^{\ast }$; reversing again is the inverse operation, and the unknot remains the unknot. Thus $-T^{\ast }$ is prime. The signatures in step2.1 show that $T$ and $-T^{\ast }$ are distinct oriented knot types, since an oriented ambient isotopy would preserve [F3]'s signature. Therefore the displayed two-prime decompositions have respective multisets $\{T,T\}$ and $\{T,-T^{\ast }\}$, which are different. By [F4]'s actual oriented Schubert uniqueness they are the prime-factor multisets of $G$ and $Q$, independent of a chosen decomposition. Together with their equal Alexander polynomials, this proves the entire original refutation. No planned matrix interface or conditional prime/invariant premise remains.