The square knot is slice despite its nonunit Alexander polynomial

The right-handed trefoil matrix is $V=\left(\begin{array}{cc}-1& 0\\ -1& -1\end{array}\right)$ by [F1] in the convention [F2]. Choose the connected-sum balls around short boundary arcs where the Seifert surfaces are standard collars, disjoint from their basis curves and small positive push-offs. The fixed gluing [F5] joins those collars. A basis cycle in one summand and its push-off lie in a ball disjoint from the other summand's curves, so the first cycle bounds a $2$-chain there away from the second; hence all cross linking entries vanish. The combined matrix is therefore $W=\mathrm{diag}(V,-V)$ by [F3].

Direct calculation gives $\det (V-t{V}^T)=\det \left(\begin{array}{cc}t-1& t\\ -1& t-1\end{array}\right)=(1-t{)}^2+t=t^2-t+1$. For the reverse-mirror block, $-V-t(-V^T)=-(V-t{V}^T)$, whose determinant equals $\det (V-t{V}^T)$ because the rank is two. Since $W-t{W}^T$ is block diagonal, its determinant is $(t^2-t+1{)}^2$. By [F2] and [F4], this is a representative of ${\mathrm{\Delta }}_S(t)$.

Expanding $(t^2-t+1{)}^2$ gives $t^4-2t^3+3t^2-2t+1$, which has nonzero coefficients in degrees $0$ and $1$ and is therefore not a monomial $\pm t^k$. Since these monomials are exactly the units of $\mathbb{Z}[t^{\pm 1}]$, the Alexander polynomial is nonunit. On the other hand, [A2] supplies AC${}_{\omega }$ from full AC, so [F7] gives a smooth ribbon disk; by [F8] this disk is locally flat. Thus $S$ is topologically slice despite its nonunit Alexander polynomial, refuting the asserted converse. For $T$, the same calculation gives ${\mathrm{\Delta }}_T=t^2-t+1\doteq ̸1$, which only means that the Alexander-one theorem does not apply.