 Full-degree irreducible Alexander polynomial implies a prime knot

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.