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Tag 077G

Chapter 10: Commutative Algebra > Section 10.55: Graded rings

Lemma 10.55.1. 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 $R$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 12682–12691 (see updates for more information).

    \begin{lemma}
    \label{lemma-integral-closure-graded}
    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 $S$.
    \end{lemma}
    
    \begin{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 \ref{lemma-integral-transitive}.
    Finally, we also conclude that $s_m = s - \sum_{i = n, \ldots, m - 1} s_i$
    is integral over $R$.
    \end{proof}

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