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Tag 00JL

10.55. Graded rings

A graded ring will be for us a ring $S$ endowed with a direct sum decomposition $S = \bigoplus_{d \geq 0} S_d$ such that $S_d \cdot S_e \subset S_{d + e}$. Note that we do not allow nonzero elements in negative degrees. The irrelevant ideal is the ideal $S_{+} = \bigoplus_{d > 0} S_d$. A graded module will be an $S$-module $M$ endowed with a direct sum decomposition $M = \bigoplus_{n\in \mathbf{Z}} M_n$ such that $S_d \cdot M_e \subset M_{d + e}$. Note that for modules we do allow nonzero elements in negative degrees. We think of $S$ as a graded $S$-module by setting $S_{-k} = (0)$ for $k > 0$. An element $x$ (resp. $f$) of $M$ (resp. $S$) is called homogeneous if $x \in M_d$ (resp. $f \in S_d$) for some $d$. A map of graded $S$-modules is a map of $S$-modules $\varphi : M \to M'$ such that $\varphi(M_d) \subset M'_d$. We do not allow maps to shift degrees. Let us denote $\text{GrHom}_0(M, N)$ the $S_0$-module of homomorphisms of graded modules from $M$ to $N$.

At this point there are the notions of graded ideal, graded quotient ring, graded submodule, graded quotient module, graded tensor product, etc. We leave it to the reader to find the relevant definitions, and lemmas. For example: A short exact sequence of graded modules is short exact in every degree.

Given a graded ring $S$, a graded $S$-module $M$ and $n \in \mathbf{Z}$ we denote $M(n)$ the graded $S$-module with $M(n)_d = M_{n + d}$. This is called the twist of $M$ by $n$. In particular we get modules $S(n)$, $n \in \mathbf{Z}$ which will play an important role in the study of projective schemes. There are some obvious functorial isomorphisms such as $(M \oplus N)(n) = M(n) \oplus N(n)$, $(M \otimes_S N)(n) = M \otimes_S N(n) = M(n) \otimes_S N$. In addition we can define a graded $S$-module structure on the $S_0$-module $$ \text{GrHom}(M, N) = \bigoplus\nolimits_{n \in \mathbf{Z}} \text{GrHom}_n(M, N), \quad \text{GrHom}_n(M, N) = \text{GrHom}_0(M, N(n)). $$ We omit the definition of the multiplication.

Let $S$ be a graded ring. Let $d \geq 1$ be an integer. We set $S^{(d)} = \bigoplus_{n \geq 0} S_{nd}$. We think of $S^{(d)}$ as a graded ring with degree $n$ summand $(S^{(d)})_n = S_{nd}$. Given a graded $S$-module $M$ we can similarly consider $M^{(d)} = \bigoplus_{n \in \mathbf{Z}} M_{nd}$ which is a graded $S^{(d)}$-module.

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 12624–12735 (see updates for more information).

    \section{Graded rings}
    \label{section-graded}
    
    \noindent
    A {\it graded ring} will be for us a ring $S$ endowed
    with a direct sum decomposition $S = \bigoplus_{d \geq 0} S_d$
    such that $S_d \cdot S_e \subset S_{d + e}$.
    Note that we do not allow nonzero elements in negative degrees.
    The {\it irrelevant ideal} is the ideal $S_{+} = \bigoplus_{d > 0} S_d$.
    A {\it graded module}
    will be an $S$-module $M$ endowed with a direct sum decomposition
    $M = \bigoplus_{n\in \mathbf{Z}} M_n$ such that $S_d \cdot M_e
    \subset M_{d + e}$. Note that for modules we do allow
    nonzero elements in negative degrees.
    We think of $S$ as a graded $S$-module by setting $S_{-k} = (0)$
    for $k > 0$. An element $x$ (resp.\ $f$) of $M$ (resp.\ $S$) is called
    {\it homogeneous}
    if $x \in M_d$ (resp.\ $f \in S_d$) for some $d$.
    A {\it map of graded $S$-modules} is a map of $S$-modules
    $\varphi : M \to M'$ such that $\varphi(M_d) \subset M'_d$.
    We do not allow maps to shift degrees. Let us denote
    $\text{GrHom}_0(M, N)$ the $S_0$-module of homomorphisms
    of graded modules from $M$ to $N$.
    
    \medskip\noindent
    At this point there are the notions of graded ideal,
    graded quotient ring, graded submodule, graded quotient module,
    graded tensor product, etc. We leave it to the reader to find the
    relevant definitions, and lemmas. For example: A short exact sequence
    of graded modules is short exact in every degree.
    
    \medskip\noindent
    Given a graded ring $S$, a graded $S$-module $M$ and $n \in \mathbf{Z}$
    we denote $M(n)$ the graded $S$-module with $M(n)_d = M_{n + d}$.
    This is called the {\it twist of $M$ by $n$}. In particular we get
    modules $S(n)$, $n \in \mathbf{Z}$ which will play an important
    role in the study of projective schemes. There are some obvious
    functorial isomorphisms such as
    $(M \oplus N)(n) = M(n) \oplus N(n)$,
    $(M \otimes_S N)(n) = M \otimes_S N(n) = M(n) \otimes_S N$.
    In addition we can define a graded $S$-module structure on
    the $S_0$-module
    $$
    \text{GrHom}(M, N) =
    \bigoplus\nolimits_{n \in \mathbf{Z}} \text{GrHom}_n(M, N),
    \quad
    \text{GrHom}_n(M, N) = \text{GrHom}_0(M, N(n)).
    $$
    We omit the definition of the multiplication.
    
    \medskip\noindent
    Let $S$ be a graded ring. Let $d \geq 1$ be an integer.
    We set $S^{(d)} = \bigoplus_{n \geq 0} S_{nd}$. We think of
    $S^{(d)}$ as a graded ring with degree $n$ summand
    $(S^{(d)})_n = S_{nd}$. Given a graded $S$-module $M$ we
    can similarly consider $M^{(d)} = \bigoplus_{n \in \mathbf{Z}} M_{nd}$
    which is a graded $S^{(d)}$-module.
    
    \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}

    Comments (1)

    Comment #675 by Keenan Kidwell on June 10, 2014 a 8:19 pm UTC

    In the proof of 077G, the bottom horizontal map in the diagram should be $r\mapsto t^{\deg(r)}r$, right?

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