Why should we study morphism of $k$-schemes instead of morphisms of schemes? I have been self-studying algebraic geometry through youtube video lecture of Prof. Richard E. BORCHERDS. This is the link of the lecture that I have question about: https://www.youtube.com/watch?v=dvKfbXhNOB0&list=PL8yHsr3EFj50Un2NpfPySgXctRQK7CLG-&index=12.
He mentioned in the lecture that we prefer to study morphisms between two schemes (from $\text{spec}(\mathbb{k}) \to \mathbb{P}^n$ for example) viewing as schemes over $\text{spec}(\mathbb{k})$ rather than morphisms between themselves. To have morphisms between, say $\text{spec}(\mathbb{k}) \to \mathbb{P}^n$, we need to have an extra commutative diagram over $\text{spec}(\mathbb{k})$. Why should we prefer that? The professor mentioned that was because we want to avoid "weird" automorphism of $\mathbb{k}$. I couldn't understand what he was trying to explain by that. Could anyone explain this a bit more details? Some concrete example would be great. Thank you!
 A: How big is $\operatorname{Aut}(\operatorname{Spec} \Bbb C)$? Well, it depends what type of maps you consider. If you consider $\Bbb C$-algebra maps, then the unique automorphism is the identity. If you consider $\Bbb R$-algebra maps, then there are two automorphisms: the identity and complex conjugation. If you consider general ring maps, there are at least $2^{2^{\aleph_0}}$ automorphisms: pick a transcendence basis for $\Bbb C$ over $\Bbb Q$, and permute the transcendence basis. That's too many automorphisms, in that it doesn't correspond to the geometric picture we want - if we're trying to generalize classical algebraic geometry, a single (closed, reduced) point shouldn't have any automorphisms.
A: Question: "Why should we prefer that? The professor mentioned that was because we want to avoid "weird" automorphism of k. I couldn't understand what he was trying to explain by that. Could anyone explain this a bit more details? Some concrete example would be great. Thank you!"
Answer: Any pair of schemes $X,Y$ have canonical maps to $S:=Spec(\mathbb{Z})$ and any map
$f: X\rightarrow Y$ is defined over $S$. Hence you may view an arbitry map of schemes as maps over the integers $\mathbb{Z}$. When working over a field $k$ you change your base scheme $S$ to $T:=Spec(k)$. There is a canonical map $\phi: \mathbb{Z}\rightarrow k$ mapping $\phi(m)=\phi(1+ \cdots + 1)=m\phi(1)=m\in k$. Hence the map $X \rightarrow S$ factors canonically as $X \rightarrow T \rightarrow S$. Hence when you are not working over a field $k$ you are working over $\mathbb{Z}$.
Example 1: If $k:=\mathbb{Q} \subseteq K:=\mathbb{Q}(i)$ you may look at two groups:
$$End_k(K)\text{ and }End_K(K).$$
Since $dim_k(K)=2$ it follows $End_k(K)=Mat(2,k)$ is the ring of $2\times 2$-matrices with coefficients in $k$ and $End_K(K)\cong K\cong k^2$. Hence when you change base field the situation changes.
Example 2: A map $\phi^*: Spec(K) \rightarrow Spec(K)$ defined over $k$
is given by an automorphism $\phi \in G(K/k)$ and there is only one map defined over $K$ - the identity map. The Galois group $G(K/k)$ enters in various formulas in algebraic number theory and arithmetic geometry, and when you change the base field, the set of morphisms changes and this relates to the Galois group.
Example 3: If $V\cong \mathbb{k}^{n+1}$ is a vector space over a field $k:=\mathbb{k}$, it follows there is a 1-1 correspondence between maps (over $k$)
$$ x: Spec(k) \rightarrow \mathbb{P}(V^*)$$
and lines $l_{x} \subseteq V$. Hence the $k$-rational points $x \in \mathbb{P}(V^*)(k)$ correspond 1-1 to lines $l_x \subseteq V$. This reflects the fact that projective space $\mathbb{P}(V^*)$ is a parameter space - it parametrize lines in $V$.
