Lemma 10.56.3. 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 10.36.6. Finally, we also conclude that s_ m = s - \sum _{i = n, \ldots , m - 1} s_ i is integral over R. \square
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