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Let $k$ be a field, and $A$ a (commutative associative unital) $k$-algebra. It is known that there is a natural isomorphism between ‘regular’ (in the sense of regular maps of varieties) actions of $k^\times$ on $\operatorname{Spec} A$ and $\mathbb{Z}$-gradings of $A$: see here, for example.

Now, suppose $k$ is algebraically closed and $A$ is finitely-generated and reduced, so that $\operatorname{Spec} A$ is an affine variety over $k$. Let $V = \operatorname{MaxSpec} A$ (and so a variety in the traditional sense), and suppose $M : V \times \mathbb{A}^1 \to V$ is a morphism of affine varieties and a multiplicative action of $k$, in the sense that $M$ satisfies the associativity condition $$M(M(p, a), a') = M(p, a a')$$ and the identity condition $$M(p, 1) = p$$ for all $p$ in $V$ and all $a$ and $a'$ in $k$, regarded as points of $\mathbb{A}^1$. If I'm not mistaken, this suffices to induce an $\mathbb{N}$-grading on $A$. Conversely, given an $\mathbb{N}$-grading on $A$ we may construct such a multiplicative action. I am interested in geometrical interpretations of properties of gradations this correspondence. For example, $M$ satisfies the absorption condition that there is a unique $\hat{p}$ in $V$ such that $$M(p, 0) = \hat{p}$$ for all $p$ in $V$ if and only if $A_0 \cong k$.

Question 1: What is the condition on the gradation which corresponds to $\hat{p}$ being the unique fixed point of $M$? (Equivalently, what is the condition which corresponds to the $k^\times$ group action being free away from $\hat{p}$?)

Question 2: What is the condition on the action which corresponds to $A$ being generated as a $A_0$-algebra by $A_1$? Neither freeness nor faithfulness seems to be the answer, since the curve $y^2 = x^3$ has an obvious multiplicative action which is both free and faithful but the induced gradation on the affine coordinate ring has nothing in degree $1$.

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