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$\newcommand{\Spec}{\mathrm{Spec}}$Recall the following result, proved in Section 2.9 of Neil Strickland's Formal schemes and formal groups, and in Lemma 1.3.2 of Eric Peterson's Formal Geometry and Bordism Operations, available also on GitHub.

Let $R$ be a ring. We have a bijection $$\left\{\begin{gathered}\text{group scheme actions}\\\mathbb{G}_{m}\times\Spec(R)\to\Spec(R)\end{gathered}\right\}\cong\left\{\text{$\mathbb{Z}$-gradings of $R$}\right\}.$$ Moreover a ring map $f\colon A\to B$ respects the grading if $\Spec(f)\colon\Spec(B)\to\Spec(A)$ is $\mathbb{G}_{m}$-equivariant.

Question. Are there generalisations of this result for $A$-gradings with $A$ a commutative monoid? In particular, do we have such a result for $(\mathbb{N},+,0)$-gradings? What about $(\mathbb{N},1,\cdot)$- or $(\mathbb{N}_{\geq1},1,\cdot)$-gradings?

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    $\begingroup$ Asking that a $\mathbf{Z}$-grading be a $\mathbf{N}$-grading just amounts to saying that the $\mathbf{G}_m = \mathrm{Spec} \ \mathbf{Z}[t^{\pm 1}]$-weights are nonnegative. One way you can say this is to consider $\mathbf{A}^1 = \mathrm{Spec} \ \mathbf{Z}[t]$ as a monoid with respect to multiplication. Then $\mathbf{A}^1$-actions correspond to nonnegative gradings. (If you consider $\mathrm{Spec}\ \mathbf{Z}[t^{-1}]$ instead, you'd get nonpositive gradings) $\endgroup$
    – skd
    Commented Sep 4, 2021 at 3:40
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    $\begingroup$ Maybe to clarify: the general picture is that if $\Lambda$ is a commutative monoid in sets, then $\mathbf{Z}[\Lambda]$ is a commutative $\mathbf{Z}$-algebra, and the diagonal on $\Lambda$ makes $\mathbf{Z}[\Lambda]$ into a cocommutative coalgebra. So $\mathrm{Spec} \ \mathbf{Z}[\Lambda]$ is a commutative group scheme, and actions of this thing correspond to $\Lambda$-gradings (which you can think of as functors $\Lambda \to \mathrm{Mod}_\mathbf{Z}$, with $\Lambda$ viewed as a discrete category). This is a categorical version of Cartier duality. $\endgroup$
    – skd
    Commented Sep 4, 2021 at 3:59
  • $\begingroup$ When the commutative semigroup $A$ has cancellation, it embeds into its group $B$ of fractions and an $A$-grading is a $B$-grading with a certain condition, which can be more or less described in terms of action. (The case of the semigroup $(\mathbf{N}_{>0},+)$, which is not a monoid, is quite important). I would have less intuition when, for instance, $A$ is non-cancelative, e.g. $A=(\{0,1\},\max)$. By the way $(\mathbf{N}_{>0},\times)\simeq \mathbf{N}^{(\mathbf{N})}$ is just a free commutative monoid on countably many generators. $\endgroup$
    – YCor
    Commented Sep 4, 2021 at 6:51
  • $\begingroup$ I think that the second comment by skd is a full answer. (At least, this is what I would have answered :).) A reference is the book by Demazure-Gabriel. $\endgroup$
    – Martin Brandenburg
    Commented Sep 4, 2021 at 7:45
  • $\begingroup$ Isn’t $Spec(Z[\Lambda])$ a monoid scheme $\endgroup$
    – Benjamin Steinberg
    Commented Sep 4, 2021 at 12:13

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