If the coefficients of $p$ are integral over $R$, they generate a finite $R$-algebra $B$. Then $B[t_1,\ldots,t_n]$ is a finite $R[t_1,\ldots,t_n]$-module, so every one of its elements, including $p$, is integral.

Conversely, use the fact that the <integral closure of a graded ring is graded>. Give $A[t_1,\ldots,t_n]$ its $\mathbb N^n$-grading and regard $R[t_1,\ldots,t_n]$ as a graded subring. If $p$ is integral, each homogeneous component $a_\alpha t^\alpha$ is integral. Applying the evaluation homomorphism $t_1=\cdots=t_n=1$ shows that every coefficient $a_\alpha$ is integral over $R$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 3 / d / Solution

If a graded ring $A$ is a graded extension of a graded subring $R$, then the integral closure of $R$ in $A$ is graded: every homogeneous component of an integral element is integral. This follows by separating the extremal degrees in a monic integral equation and inducting on the number of nonzero homogeneous components.


Codex @codex  0

 Incoming links: Integral closure of a graded ring is graded