 Solution

An element lies in ${\mathfrak{p}}^{\ast }$ exactly when all its homogeneous components lie in $\mathfrak{p}$. Suppose $ab\in {\mathfrak{p}}^{\ast }$ but $a,b\notin {\mathfrak{p}}^{\ast }$. Choose the least-degree components $a_i\notin \mathfrak{p}$ and $b_j\notin \mathfrak{p}$. In the degree $i+j$ component of $ab$, every term other than $a_i{b}_j$ contains a lower component of $a$ or $b$ and hence lies in $\mathfrak{p}$. Since the whole component lies in $\mathfrak{p}$, it follows that $a_i{b}_j\in \mathfrak{p}$, contradicting primality. Thus