Functor of points of projective varieties (e.g., elliptic curve) over any commutative ring Assume that $V \subset \mathbb P^n$ is a closed subvariety of a projective space over some field $k$. For instance $n=2, k = \mathbb F_5$ and $V : Y^2 Z = X^3 + Z^3$ is an elliptic curve.
For any $k$-algebra $R$, we should have an embedding $V(R) \subset \mathbb P^n(R)$. Now, the $R$-rational points of $\mathbb P^n$ are described here (see also here and there) as equivalence classes of tuples $(L, (s_0, ..., s_n))$ where $L$ is a line bundle over $\mathrm{Spec}(R)$ and $s_i \in L(R)$ are global sections (i.e., elements of the $R$-module $L(R)$) that "generate" $L$.

My question is:
how does $V(R)$ look like inside $\mathbb P^n(R)$ under this identification?

I do not see a way to multiply the sections $s_i$ since they are elements of a module and not of a ring (unless $L = O_{\mathrm{Spec}(R)}$ is the trivial line bundle, which is automatic if $R$ has trivial Picard group).
 A: I'm not very well versed in working over non-algebraically closed fields, so I'm not sure everything I'm saying is true also in this case. So I will assume $k$ to be algebraically closed for simplicity, but I think it should also hold in general.
There is a way of multiplying such sections, by using tensor products. Namely, given a line bundle $\mathcal{L}$, we can associate to it the graded ring $R(X,\mathcal{L}):=\bigoplus_{n\geq 0}\Gamma(X,\mathcal{L}^{\otimes n})$ (where $\mathcal{L}^{\otimes 0}:=\mathcal{O}_X$). Multiplication of $s\in\mathcal{L}^{\otimes m}$ and $t\in\mathcal{L}^{\otimes n}$ is defined as $s\otimes t\in\mathcal{L}^{\otimes(m+n)}$.
So let us consider the description in your first link. A morphism $\operatorname{Spec}(R)\to\mathbb{P}^n$ corresponds to a line bundle $\mathcal{L}$ on $\operatorname{Spec}R$ together with global sections $s_0,\ldots,s_n$ which globally generate $\mathcal{L}$ (i.e. don’t vanish simultaneously). Intuitively (as opposed to precisely), for some closed point $x\in \operatorname{Spec}R$ for which $(s_i)_x\notin \mathfrak{m}_x\mathcal{L}_x$ (i.e. $s_i$ doesn't vanish at $x$), while it doesn't make sense to "evaluate" the sections at $x$, there are unique scalars $\lambda_0,\ldots,\lambda_n\in k$ such that $(s_j)_x=\lambda_j (s_i)_x$ (in particular $\lambda_i=1$), and the morphism corresponding to $(\mathcal{L},s_0,\ldots,s_n)$ will send $x$ to $[\lambda_0:\ldots:\lambda_n]$.
Still intuitively, we consider your example where $V\subseteq\mathbb{P}^2$ is defined by $Y^2Z=X^3+Y^3$. Take a line bundle $\mathcal{L}$ on $\operatorname{Spec}R$ and generating sections $s_0,s_1,s_2$. If now $s_1^2s_2=s_0^3+s_1^3$ inside $R(\operatorname{Spec} R,\mathcal{L})$, then the morphism determined by $(\mathcal{L},s_0,s_1,s_2)$ has image inside $V$. Indeed, let say that $s_0$ doesn't vanish at $x$, and that $(s_1)_x=\lambda_1(s_0)_x$, $(s_2)_x=\lambda_2(s_0)_x$ for some scalars $\lambda_1,\lambda_2$. Then as $s_1^2s_2=s_0^3+s_1^3$ inside $R(\operatorname{Spec} R,\mathcal{L})$, we then obtain that $\lambda_1^2\lambda_2=1^3+\lambda_1^3$.
One can actually make this formal: if a morphism $\phi:X\to\mathbb{P}^n$ is determined by $(\mathcal{L},s_0,\ldots,s_n)$, then this gives a morphism of graded $k$-algebras $\phi^\flat:R(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(1))\to R(X,\mathcal{L})$ via pullback. By Hartshorne II.5.13, we have $R(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(1))\cong k[x_0,\ldots,x_n]$, and $\phi^{\flat}$ maps $x_i$ to $s_i$. Then we actually have $\overline{\phi(X)}=V(\ker\phi^{\flat})$.
