Consider the group scheme $\mathbb{G}_m = \mathrm{Spec}(\mathbb{Z}[t, t^{-1}])$. An action of $\mathbb{G}_m$ on an affine scheme $X = \mathrm{Spec}(A)$ is a morphism of schemes $$ \sigma: \mathbb{G}_m \times X \to X $$ such that for every scheme $T$, $\sigma_T : \mathbb{G}_m(T) \times X(T) \to X(T)$ is a group action in the standard definition. I want to show that the action $\sigma$ is equivalent to a $\mathbb{Z}$-grading on $A$. I've been trying to do this for a little while but I am stuck. How can I prove/think about this?
The closest I've got to solving this is by using the idea of considering $A$ as a $\mathbb{Z}[t, t^{-1}]$-comodule using the ring homomorphism $$ \Delta: A \to A \otimes \mathbb{Z}[t, t^{-1}] \cong A[t, t^{-1}] $$ dual to the action $\sigma$. This was taking from example 1.6.6 in Hida's book on geometric modular forms, which refers to Jantzen's book on Reps of Algebraic Groups. However, I'm not comfortable with the idea of comodules, including why this is a comodule map. However, I think I see why, if this is a comodule map and satisfies the requesite properties mentioned in Jantzen, then we get a grading. Given a grading, there seems to be an obvious way to define a homomorphism $\Delta$ taking $a \mapsto \sum_{n \in \mathbb{Z}} p_n(a) \otimes t^n$ where $p_n$ is the projection map onto the $n^{th}$ graded piece. Is it true that this gives a group action? Again I'm not that comfortable with this equivalence between group actions and $G$-modules and $\mathbb{Z}[t, t^{-1}]$-comodules. If the above description is the best way to think of this then I will try to understand these better, but I wanted an outside opinion on how to prove/think about this.