Lemma 39.12.4. Let $a : \mathbf{G}_ m \times X \to X$ be an action on an affine scheme. Then $X$ is the spectrum of a $\mathbf{Z}$-graded ring and the action is as in Example 39.12.3.

Proof. Let $f \in A = \Gamma (X, \mathcal{O}_ X)$. Then we can write

$a^\sharp (f) = \sum \nolimits _{n \in \mathbf{Z}} f_ n \otimes x^ n \quad \text{in}\quad A \otimes \mathbf{Z}[x, x^{-1}] = \Gamma (\mathbf{G}_ m \times X, \mathcal{O}_{\mathbf{G}_ m \times X})$

as a finite sum with $f_ n$ in $A$ uniquely determined. Thus we obtain maps $A \to A$, $f \mapsto f_ n$. Since $a$ is an action, if we evaluate at $x = 1$, we see $f = \sum f_ n$. Since $a$ is an action we find that

$\sum (f_ n)_ m \otimes x^ m \otimes x^ n = \sum f_ n x^ n \otimes x^ n$

(compare with computation in Example 39.12.3). Thus $(f_ n)_ m = 0$ if $n \not= m$ and $(f_ n)_ n = f_ n$. Thus if we set

$A_ n = \{ f \in A \mid f_ n = f\}$

then we get $A = \sum A_ n$. On the other hand, the sum has to be direct since $f = 0$ implies $f_ n = 0$ in the situation above. $\square$

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