Let $X$ be a scheme over an algebraically closed field $k$. A cone over $X$ is an $X$-scheme $C\to X$ realized as the global Spec of a graded $\mathcal O_X$-algebra $S^\bullet=\bigoplus_{n\geq 0}S^n$, such that $\mathcal O_X\to S^0$ is an isomorphism, $S^1$ is coherent and generates $S$ as an algebra.
Such a cone carries a natural "$0$-section" $X\to C$ (coming from the morphism $S\to S_0\cong \mathcal O_X$), and a compatible $\mathbb A^1$-action $C\times\mathbb A^1\to C$ (coming from the augmentation $\epsilon:S\to S^\bullet[z]$, where $S^\bullet[z]=\bigoplus_{n\geq 0}\bigl(S^n\oplus S^{n-1}z\oplus\dots\oplus S^1z^{n-1}\oplus S^0z^n\bigr)$ is another $\mathcal O_X$-algebra satisfying our requirements. I think the augmentation $\epsilon$ is defined by sending $s_i\in S^i$ to $s_iz^i$; this should make $\epsilon$ into a graded morphism).
Question. Let $C$ be the global Spec of an $\mathcal O_X$-algebra $S^\bullet$ such that $S^0\cong \mathcal O_X$ and $S^\bullet$ is generated in degree $1$ by its coherent part $S^1$. Could you please provide a proof, or a reference, of the fact that to give a $0$-section and a compatible $\mathbb A^1$-action is the same as to choose a grading on $S$ (i.e. to make $C$ into a cone over $X$)?
If we have the grading, it is not so hard to induce the $0$-section and the compatible $\mathbb A^1$-action. I roughly described this two paragraphs ago. But I still miss the inverse construction.
Thank you all in advance!