
Lemma 10.55.3. Let $R \to S$ be a homomorphism of graded rings. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then

$S' = \bigoplus \nolimits _{d \geq 0} S' \cap S_ d,$

i.e., $S'$ is a graded $R$-subalgebra of $S$.

Proof. We have to show the following: If $s = s_ n + s_{n + 1} + \ldots + s_ m \in S'$, then each homogeneous part $s_ j \in S'$. We will prove this by induction on $m - n$ over all homomorphisms $R \to S$ of graded rings. First note that it is immediate that $s_0$ is integral over $R_0$ (hence over $R$) as there is a ring map $S \to S_0$ compatible with the ring map $R \to R_0$. Thus, after replacing $s$ by $s - s_0$, we may assume $n > 0$. Consider the extension of graded rings $R[t, t^{-1}] \to S[t, t^{-1}]$ where $t$ has degree $0$. There is a commutative diagram

$\xymatrix{ S[t, t^{-1}] \ar[rr]_{s \mapsto t^{\deg (s)}s} & & S[t, t^{-1}] \\ R[t, t^{-1}] \ar[u] \ar[rr]^{r \mapsto t^{\deg (r)}r} & & R[t, t^{-1}] \ar[u] }$

where the horizontal maps are ring automorphisms. Hence the integral closure $C$ of $S[t, t^{-1}]$ over $R[t, t^{-1}]$ maps into itself. Thus we see that

$t^ m(s_ n + s_{n + 1} + \ldots + s_ m) - (t^ ns_ n + t^{n + 1}s_{n + 1} + \ldots + t^ ms_ m) \in C$

which implies by induction hypothesis that each $(t^ m - t^ i)s_ i \in C$ for $i = n, \ldots , m - 1$. Note that for any ring $A$ and $m > i \geq n > 0$ we have $A[t, t^{-1}]/(t^ m - t^ i - 1) \cong A[t]/(t^ m - t^ i - 1) \supset A$ because $t(t^{m - 1} - t^{i - 1}) = 1$ in $A[t]/(t^ m - t^ i - 1)$. Since $t^ m - t^ i$ maps to $1$ we see the image of $s_ i$ in the ring $S[t]/(t^ m - t^ i - 1)$ is integral over $R[t]/(t^ m - t^ i - 1)$ for $i = n, \ldots , m - 1$. Since $R \to R[t]/(t^ m - t^ i - 1)$ is finite we see that $s_ i$ is integral over $R$ by transitivity, see Lemma 10.35.6. Finally, we also conclude that $s_ m = s - \sum _{i = n, \ldots , m - 1} s_ i$ is integral over $R$. $\square$

There are also:

• 1 comment(s) on Section 10.55: Graded rings

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).