# The Stacks Project

## Tag 077G

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}
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}

There are no comments yet for this tag.

There is also 1 comment on Section 10.55: Commutative Algebra.

## Add a comment on tag 077G

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 lower-right corner).