L-valued pt of Projective Space versus Classical Points

In the previous post, we constructed $\mathbb{P}^n(k)$ by gluing affine schemes. Now we will show that if $k$ is algebraically closed, then the closed points of $\mathbb{P}^n(k)$ are in bijective correspondence with points of the classic projective space which we will denote $P^n(k)$. More generally, for any field $k$ and any extension $L$ of $k$, we have that the $K$-points of $\mathbb{P}^n(k)$, i.e. $\hom_k(\text{Spec} L, \mathbb{P}^n(k))$ is in bijective correspondence with $P^n(L)$.

$L$-valued points of $\mathbb{P}^n(k)$ are Classical Points

 Let $\overline{x}: \text{Spec} L \to \mathbb{P}^n$ be an $L$-valued point of $\mathbb{P}^n(k)$ with image $x$. Then $x$ lies in one of the pieces $X_i = \text{Spec} A_i$ where $$A_i = k\left[\frac{t_0}{t_i}, \ldots, \frac{t_n}{t_i}\right].$$ So $\overline{x}$ corresponds to a morphism $\sigma_i: A_i \to L$, so we can associate to it the point $$\left[\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: 1 :\ldots: \sigma_i\left(\frac{t_n}{t_i}\right)\right]$$ of $P^n(L)$.

Well-defined.

 This assignment is well-defined. Indeed, if $x \in X_j$ also, then $\overline{x}$ factors through $X_i \cap X_j$. This means that $\sigma_i$ and $\sigma_j$ \emph{extends} to a homomorphism $$\sigma_{ij}: A_{ij} = k\left[t_0, \ldots, t_n, \frac{1}{t_i}, \frac{1}{t_j}\right] \to L.$$
Then we can write $$\left[\sigma_j\left(\frac{t_0}{t_j}\right) : \ldots: \sigma_j\left(\frac{t_n}{t_j}\right)\right] $$ as $$\left[\sigma_{ij}\left( \frac{t_i}{t_j}\right)\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: \sigma_{ij}\left( \frac{t_i}{t_j}\right)\sigma_i\left(\frac{t_n}{t_i}\right)\right]$$ which is the same as $$\left[\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: \sigma_i\left(\frac{t_n}{t_i}\right)\right].$$

Bijection

For each choice of element of $X_i$ we get a unique morphism $\sigma_i: A_i \to L$ and hence from $\text{Spec} L \to X_i \to X$. 

If $k= \overline{k}$ then closed points of  $\mathbb{P}^n(k)$ are Classical Points

It suffices to show that for $k$ algebraically closed, and $X$ any scheme over $k$ i.e. with a canonical morphism $X \to \text{Spec }k$.
$$ k\text{-points of $X$} \leftrightarrow \text{ closed points of $X$}.$$

Notice that for every point $x \in X$, restricting the canonical morphism to an affine piece piece $\text{Spec} A$ over $X$ gives a homomorphism $k \to A$. Thus this induces a map $k \to \kappa(x)$

Indeed let $\overline{x}: \text{Spec} k \to X$ be a $k$-point with image $x$. Then, as this is a morphism of schemes, we have an induced map on function fields $\kappa(x) \to k$  compatible with the canonical morphism $k \to \kappa(x)$. So we must have $\kappa(x) \cong k$ canonically. On the other hand $\kappa(x) = \text{Frac}(A/\mathfrak{p}_x) \supset A/\mathfrak{p}_x \supset k$

Do I need $A$ to be finitely generated?