In Exercise 6.6.S(b) in Vakil’s FOAG, it is stated that a $\mathbb{Z}$-valued grading on a ring $A$ is equivalent to a left $\mathbb{G}_m$-action on $\operatorname{Spec} A$. I think I see how one direction goes: if $A$ has a $\mathbb{Z}$-grading, then there is a canonical ring morphism $A \to A[x,x^{-1}]$ sending $a \in A$ to the polynomial $\sum_{n \in \mathbb{Z}} a_n x^n$, where each $a_n$ is homogeneous obtaining by decomposing $a$ into the sum of homogeneous elements. Using the fact that $A[x,x^{-1}] \cong \mathbb{Z}[x,x^{-1}] \otimes_\mathbb{Z} A$, we obtain a scheme morphism $\mathbb{G}_m \times \operatorname{Spec} A \to \operatorname{Spec} A$. What I am struggling in is the opposite direction.
Now suppose we have a left $\mathbb{G}_m$-action map $a: \mathbb{G}_m \times \operatorname{Spec} A \to \operatorname{Spec} A$. The equivalence of categories $\mathbf{CRing}^\mathrm{op} \simeq \mathbf{AffSch}$ turns commutative diagrams of a (left) group object action (stating the associtaivity an identity for the action) into the commutative diagrams in $\mathbf{CRing}$:
And I am not sure how to get a $\mathbb{Z}$-grading on $A$ from these two diagrams. The exercise also says there are two ways to define the grading, but at this stage I still don't see where the "splitting point" comes from.
I would like to see at least one of the ways to define the grading (of course both is better). And here is a further natural question. Suppose $A$ is a graded ring. Then the first paragraph gives a construction for a (possible) left $\mathbb{G}_m$-action of $\operatorname{Spec} A$. Then if we construct a grading on $A$ via the action, do we get original grading back (using either way? or only one of the ways? or not at all?) Just another question out of curiousity: how is this (or these) interpretation(s) of grading is (are) used in the literature?
Thanks for helping.