 Solution

Pass to the graded domain $R=S/{\mathfrak{p}}^{\ast }$ and let $\overline{\mathfrak{p}}=\mathfrak{p}/{\mathfrak{p}}^{\ast }$. This is a nonzero prime containing no nonzero homogeneous element. Localize at the multiplicative set $U$ of all nonzero homogeneous elements. Every nonzero homogeneous element of $U^{-1}R$ is a unit; its nonzero graded pieces are one-dimensional over the degree-zero field, so after reindexing degrees this localization is a Laurent polynomial ring $K[t,t^{-1}]$. The extended prime $U^{-1}\overline{\mathfrak{p}}$ is therefore a nonzero prime of height one.

Any prime strictly between $0$ and $\overline{\mathfrak{p}}$ would remain a nonzero prime strictly below it after localization, impossible in $K[t,t^{-1}]$. Contracting back proves that no prime lies strictly between ${\mathfrak{p}}^{\ast }$ and $\mathfrak{p}$. The graded height theorem, equivalently the same localization argument applied to saturated chains, then gives