 Solution

First, $\sqrt{{\mathfrak{q}}^{\ast }}=\mathfrak{p}$. Indeed, each homogeneous element of $\mathfrak{p}=\sqrt{\mathfrak{q}}$ has a power in $\mathfrak{q}$, and that power is homogeneous and hence lies in ${\mathfrak{q}}^{\ast }$; finite homogeneous generators of the Noetherian ideal $\mathfrak{p}$ give the assertion for every element.

Now suppose $ab\in {\mathfrak{q}}^{\ast }$ and $a\notin \mathfrak{p}$. Choose the least homogeneous component $a_i\notin \mathfrak{p}$. If $b\notin {\mathfrak{q}}^{\ast }$, choose the least component $b_j\notin \mathfrak{q}$. The degree $i+j$ component of $ab$ differs from $a_i{b}_j$ by terms in $\mathfrak{q}$. Since it lies in $\mathfrak{q}$, we get $a_i{b}_j\in \mathfrak{q}$. The $\mathfrak{p}$-primary property and $a_i\notin \mathfrak{p}$ imply $b_j\in \mathfrak{q}$, a contradiction. Hence $b\in {\mathfrak{q}}^{\ast }$, proving that