Suppose $X$ is a scheme and $\mathscr{L}$ an invertible sheaf on it. Given a sequence $\left[s_0, s_1, s_2, \ldots\right]$ of global sections on it that doesn’t have common zeros, does it induce a morphism from $X$ to $\Bbb{P}^\infty_k:=\operatorname{Proj}k[x_0, x_1, x_2, \ldots]$? And does every morphism from $X$ to $\Bbb{P}^\infty_k$ have such a form?
The question is motivated from the case for $\Bbb{P}^n_k$, in which the result hold from reference like Stacks#01ND. I also posted a question earlier here, which also motivates this for a concrete coordinate example (although it is kind of duplicate, but anyway I post it as a new question here rather than change the old question since it’s some kind standalone too).
Any help would be greatly appreciated. Thank you in advance :D
Yes, this is true, and can be found on the Stacks Project as tag 01NB, which for a graded ring $S$, a scheme $Y$, an invertible sheaf $\mathcal{L}$ on $Y$, and a graded ring homomorphism $\psi:S^{(d)}\to\Gamma_*(Y,\mathcal{L})$ such that $\mathcal{L}$ is generated by the sections in the image gives a unique map $\varphi: Y \to \operatorname{Proj} S$ such that $\varphi^*(\mathcal{O}(d))\cong \mathcal{L}$ in a fashion compatible with $\psi$. The proof is substantially similar to the standard one.
The rest of the section which this lemma is found in shows that this really does everything you'd need it to, in terms of representing functors, etc.