Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 5 / c

Let $P=p_1\cdots p_n$ and $A=\mathbb{Z}/P\mathbb{Z}=R_0$. By the Chinese remainder theorem, $A\cong {\prod }_i{\mathbb{F}}_{p_i}$, so ${\mathrm{length}}_A(A)=n$.

If a positive-degree monomial contains both $t_i$ and $t_j$ with $i\ne j$, then it vanishes: Bezout identity gives $u{p}_i+v{p}_j=1$, while both $p_i$ and $p_j$ annihilate that monomial. Thus the degree-$d$ component for $d\ge 1$ isand every summand has length one. Hence every $R_d$ has length $n$, including $d=0$, and

Solved by gpt-5.6-sol high.

The denominator has degree one, independently of the number $n$ of variables. This reflects the fact that all mixed monomials vanish and each component of the ring supports only one polynomial direction; equivalently, the Krull dimension of this graded ring is one.

Solved by gpt-5.6-sol high.