= Full-degree irreducible Alexander polynomial implies a prime knot
{title2=$\operatorname{br}\Delta_K=2g_s(K),\quad\Delta_K\text{ irreducible}\ \Longrightarrow\ K\text{ prime}$}
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.
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