Let $k$ be a field and consider an arbitrary point $\alpha = (\alpha_0, \dots, \alpha_n) \in \mathbf{P}(k^{n+1})$. Then there is a bijection $\rho: \mathbf{P}(k^{n+1}) \to \mathbf{P}_k^n (k)$, the latter denoting the set of rational points of $\mathbf{P}_k^n$, where $\rho (\alpha)$ is the ideal generated by $\alpha_j T_i - \alpha_i T_j, 0\le i,j \le n$.
I am interested in only surjectivity.
Fact: The set of rational points for a scheme $X$ over a field $k$, $X(k)$ can be identified with $x \in X$ such that $k(x) = k$, where $k(x)$ is the residue field at x.
In the proof, we will use the above fact without proof or mention.
Let $x \in \mathbf{P}_k^n (k)$ and WLOG, $x \in D_+ (T_0)$. Let $\alpha_i$ be the image of $T_0^{-1}T_i \in \mathcal{O}(D_+(T_0))$ in $k = k(x)$ and set $\alpha = (\alpha_0, \dots, \alpha_n)$. Then $\rho (\alpha) = x$.
- Is WLOG because the $x \in D_+ (T_i)$ for some $i$? Why should this be the case?
Edit: I wonder if we can say that since $x$ is a rational point, in particular, it is a point in $\mathbf{P}^n = \textrm{Proj } k[T_i]$, then it cannot contain all of the $T_i$ therefore, there must be some $i$ which $x \in D_+ (T_i)$?
- $\mathcal{O}(D_+(T_0)) = k[T_0^{-1}T_i]$. So if we say that $x$ corresponded to some $p$, then we have a natural injection $k[T_0^{-1}T_i] \to k[T_i]_{(p)} / pk[T_i]_{(p)}$ since $p \in D_+(T_0)$ so just sending $T_0^{-1}T_i$ to "itself" is a well-defined map and so we can take this to be $\alpha_i$?
- I am not sure how to now see that $\rho (\alpha) = x$.
Using the above notation, it seems that $\rho (\alpha) = \alpha_j T_i + \alpha_i T_j = \overline{T_0^{-1}T_j}T_i + \overline{T_0^{-1}T_i}T_j$ but I don't see how this should relate back to $x$.