When considering a finite-type scheme as a ringed space, is it enough to look at its $k$-points? I am reading a set of notes by Michel Brion about automorphism groups of projective varieties.
The following claim appears in the proof of a theorem stating that if G is a connected group scheme, $X$ a $G$-scheme, and $\pi:X \to Y$ a proper morphism such that $\pi_*(\mathcal{O}_X)=\mathcal{O}_Y$, then there is a unique $G$-action on $Y$ such that $\pi$ is $G$-equivariant.
We will consider a finite-type scheme $Z$, as a ringed space $(Z(k), \mathcal{O}_Z)$, where the set $Z(k)$ is equipped with the Zariski topology, this makes sense as $Z$ is of finite type.
What about considering $Z$ as a ringed space makes it ok to consider only its $k$-points, and why is it necessary that $Z$ be of finite-type?
When $k$ is algebraically closed and $Z,Z'$ are of finite type over $k$:

the data of $(Z(k),\mathcal{O}_Z|_{Z(k)})$ is enough to recover $Z$, and
given a morphism $\varphi:(Z(k),\mathcal{O}_Z|_{Z(k)})\to (Z'(k),\mathcal{O}_{Z'}|_{Z'(k)})$, there is a unique morphism $\psi:Z\to Z'$ which has $\varphi$ as its restriction to the $k$-points. (Additionally, no two distinct morphisms $Z\to Z'$ restrict to the same morphism $Z(k)\to Z'(k)$.)

The recipe for recovering $Z$ from $Z(k)$ is straightforward: we need to freely adjoin generic points, which we can do by applying a nifty functor. Define a functor $t(-):Top\to Top$ which associates to any topological space $X$ the set of its irreducible closed subsets with the topology given by declaring that for every closed $Y\subset X$ we have that $t(Y)\subset t(X)$ is a closed set. This functor acts on a morphism $f:X\to X'$ by sending an irreducible closed subset $Y\subset X$ to $\overline{f(Y)}\subset X'$. For any topological space $X$, there is an induced morphism $\alpha:X\to t(X)$ by sending $x\mapsto \overline{\{x\}}$, and this map induces a bijection between open subsets of $X$ and $t(X)$.
To verify that $(t(Z(k)),\alpha_*(\mathcal{O}_Z|_{Z(k)}))\cong Z$, we'll need to make some observations about $Z(k)$. First, $Z(k)$ can be identified with the set of closed points of $Z$: by Zariski's lemma, the closed points of a scheme of finite type over a field $F$ are exactly those points which have as their residue field a finite extension of $F$. If $F$ is algebraically closed, it has no nontrivial finite extensions, so the closed points of $Z$ are exactly $Z(k)$, and they have the induced topology from $Z$. Next, for any scheme of finite type over a field, the closed points are very dense: they're dense in every nonempty closed subset.
Now we may define a map of topological spaces $t(Z(k))\to Z$ by sending any irreducible closed subset $Y\subset Z(k)$ to the generic point of its closure in $Z$. It's not difficult to see that this is a homeomorphism by our previous observations, and as $\alpha: Z(k)\to t(Z(k))$ gives a bijection on open sets, we have that $\alpha_*(\mathcal{O}_Z|_{Z(k)})\cong \mathcal{O}_Z$.
This rather quickly settles the second part, too: given $\varphi$, we can let $\psi$ be the morphism $(t(\varphi),\alpha_*\varphi^\sharp)$, and checking uniqueness is not difficult.
Essentially what we've proven here is that there's an equivalence of categories between schemes of finite type over an algebraically closed field $k$ and the $k$-points of schemes of finite type over an algebraically closed field equipped with the restriction of the structure sheaf. The upshot of this for you is that if you can construct the map in the statement you wish to prove in this restricted setting, you'll automatically get the map of schemes you want.

Both algebraically closed and finite type are necessary conditions: if either of these are violated, there are $k$-schemes which cannot be distinguished from the empty $k$-scheme by looking at their $k$-points, and our proof above breaks. In the non-algebraically closed case, consider the spectrum of the algebraic closure, and in the non-finite type case, consider the spectrum of the transcendental extension $k(x)$. In both cases, a $k$-point is equivalent to a series of $k$-algebra map $k\to F\to k$ which is the identity. But this can't happen: the map $F\to k$ must have nontrivial kernel by dimension reasons, which means that $F\to k$ is the zero map because $F$ is a field.
Question: &quot;When considering a finite-type scheme as a ringed space, is it enough to look at its k-points?&quot;
Answer: If $k$ is a field and $A:=k[x,y]/(y-x)^{n+1}$, $S:=Spec(A)$ it follows
$$S(k):=Hom_{k-alg}(A,k)$$
is the set of maps of $k$-algebras $\phi: A \rightarrow k$. The map $\phi$ is determined by its values $\phi(x)=a, \phi(y)=b$ with $(b-a)^{n+1}=0$, and since $k$ is a field it follows $b=a$. Hence $S$ has the same $k$-rational points as the affine line $\mathbb{A}^1_k:=Spec(k[t])$. There is no isomorphism $S \cong \mathbb{A}^1_k$. Hence $S$ is not determined by $S(k)$
For simplicity, Let $S:=Spec(A)$ be an integral affine scheme of finite type over an algebraically closed field $k$, with $A:=k[x_1,..,x_n]/I$ where $I$ is a prime ideal. You may consider the set $S(k)$ of $k$-rational points of $S$ and embed this set into $k^n$ - in the sense of Hartshorne, Chapter I, it follows $S(k)\subseteq k^n$ is an algebraic variety, and $I(S(k))=I$ by the Nullstellensatz. Hence you recover the ring $A$ from $S(k)$: There is an isomorphism of $k$-algebras
$$R \cong S_k:=k[x_1,..,x_n]/I(S(k))$$
and
$$S\cong Spec(S_k).$$
Hence if $k$ is algebraically closed and $S$ is an affine integral scheme of finite type over $k$  with the above construction you recover $S$ from $S(k)$. There are group schemes that are non-reduced.
Question: &quot;What is the underlying topological space of Z if not Z(k)?&quot;
Answer: Is $A:=\mathbb{R}[x]$ and $S:=Spec(A)$ it follows the underlying topological space of $S$ is the set of prime ideals in $A$. The prime ideals $I$ in $A$ have - by the nullstellensatz - a residue field that is a finite extension of $k:=\mathbb{R}$. Hence they are on the form
$$I:=(x-r)$$
for a real number $r$ or $I:=(p_z(x))$
where $p_z(x):=(x-z)(x-\overline{z}):=x^2-2ax+a^2+b^2$ where $b\in \mathbb{R}^*$. Hence there is no equality between the topological space of $S$ and the set $S(k) \cong k$.
Question: &quot;We will consider a finite-type scheme $Z$, as a ringed space $(Z(k),O_Z)$, where the set $Z(k)$ is equipped with the Zariski topology, this makes sense as $Z$ is of finite type.&quot;
Answer: Note: You must make sense of $\mathcal{O}_Z$ - $\mathcal{O}_Z$ is not a sheaf of rings on $Z(k) \subseteq Z$ which is a topological subspace.
A finitely generated $k$-algebra $A$ is a Hilbert-Jacobson ring. In such a ring any prime ideal is the intersection ot the maximal ideals containing it. Let $X:=Spec(A)$ and let $X^m \subseteq X$ be the set of closed points in $X$ with the induced topology. Let $i: X^m \rightarrow X$ be the inclusion map. The map $i$ is continuous and you may take the topological inverse image $\mathcal{O}_{X^m}:=i^{-1}(\mathcal{O}_X)$ to get a ringed topological space $F(A):=(X^m, \mathcal{O}_{X^m})$. This defines in a functorial way a ringed topological space $F(A)$ to any finitely generated $k$-algebra, and for any morphism $\phi: A \rightarrow B$ you get a canonical map of ringed spaces
$$F(\phi): F(B) \rightarrow F(A).$$
With this construction any map of ringed spaces $f: F(B) \rightarrow F(A)$
comes from a map of rings $\phi: A \rightarrow B$ and if $F(\phi)=F(\psi)$ it follows $\phi=\psi$. If $k$ is algebraically closed it follows $X^m=X(k)$ is the space of $k$-rational points of $X$.
More generally if $X$ is of finite type over a field or Dedekind domain $k$, you let $X^m \subseteq X$ be the topological subspace of closed points of $X$ and define $F(X):=(X^m, \mathcal{O}_{X^m})$ similarly. You get to any scheme $X$ of finite type over $k$ a ringed space $F(X)$ which is functorial in $X$.
Any morphism $F(X) \rightarrow F(Y)$ comes from a map of schemes $X \rightarrow Y$ and if $F(\phi)=F(\psi)$ it follows $\phi=\psi$. Hence the ringed space $F(X)$ - which only depends on the closed points of $X$ - captures all properties of the scheme $X$. If $k$ is an algebraically closed field it follows $F(X)=X(k)$ is the space of $k$-rational points of $X$. This construction works over any Dedekind domain. The construction is &quot;straightforward&quot; and you avoid too many abstract &quot;category theoretical&quot; results that may lead to set theoretical problems: Since $X$ is a set it follows $X^m$ is also a set since it is a subset of $X$.
You find an explanation of this construction at the following link:
https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961