Lemma 39.12.5. Let $A$ be a graded ring. Let $X = \mathop{\mathrm{Spec}}(A)$ with action $a : \mathbf{G}_ m \times X \to X$ as in Example 39.12.3. Let $\mathcal{F}$ be a $\mathbf{G}_ m$-equivariant quasi-coherent $\mathcal{O}_ X$-module. Then $M = \Gamma (X, \mathcal{F})$ has a canonical grading such that it is a graded $A$-module and such that the isomorphism $\widetilde{M} \to \mathcal{F}$ (Schemes, Lemma 26.7.4) is an isomorphism of $\mathbf{G}_ m$-equivariant modules where the $\mathbf{G}_ m$-equivariant structure on $\widetilde{M}$ is the one from Example 39.12.3.

Proof. You can either prove this by repeating the arguments of Lemma 39.12.4 for the module $M$. Alternatively, you can consider the scheme $(X', \mathcal{O}_{X'}) = (X, \mathcal{O}_ X \oplus \mathcal{F})$ where $\mathcal{F}$ is viewed as an ideal of square zero. There is a natural action $a' : \mathbf{G}_ m \times X' \to X'$ defined using the action on $X$ and on $\mathcal{F}$. Then apply Lemma 39.12.4 to $X'$ and conclude. (The nice thing about this argument is that it immediately shows that the grading on $A$ and $M$ are compatible, i.e., that $M$ is a graded $A$-module.) Details omitted. $\square$

Comment #3775 by Herman Rohrbach on

Typo: in the last sentence of the statement of the lemma, "onde" should be "one".

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