Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 101 / 3

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

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

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