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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 $R$. $\square$

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