In scheme-theory, "The projective $n$-dimensional space over $k$" is defined as $\mathbb P^n_k:=\text{Proj}(k[T_0,\ldots,T_n])$. Moreover $\mathbb P^n_k$ is endowed with a structure of variery over $k$ thans to a morphism $f:\mathbb P^n_k\longrightarrow\text{Spec}\, k$.
Is there a canonical choice of $f$?
Formally if I choose a another morphism $g$ from $\mathbb P^n_k$ onto $\text{Spec} \, k$, then I get a "different" projective space (as projective variety)!
Example:
Suppose an imaginary theorem which says that there are only $100$ curves (up to isomorphism of varieties) with a morphism (ie. morphism of $k$-schemes) onto $\mathbb P^1_k$. Then can I use any structure of $k$-scheme on $\mathbb P^1_k$ to construct one of such morphisms? To be more precise If $\alpha:X\longrightarrow\mathbb P^1_k$ and $\beta:Y\longrightarrow(\mathbb P^1_k)'$ are two morphisms of curves where $\mathbb P^1_k$ and $(\mathbb P^1_k)'$ differ only in the structural morphim, then can I say that $\alpha$ and $\beta$ are two of the $100$ morphisms described in the imaginary theorem?
Yes, there is a canonical map $\: \mathbb P^n_k \to \operatorname {Spec(k)}$ for $k$ an arbitrary ring but strangely, even if $k$ is a field, it cannot be defined in an elegant functorial way but is defined by a rather clumsy glueing procedure. More precisely:
For every homogeneous polynomial of positive degree $f\in k[T_0,\cdots ,T_n]$ we get the open subset $D_+(f)=\operatorname {Spec(k[T_0,\cdots ,T_n]_{(f)})}\subset \mathbb P^n_k$ and a scheme morphism $D_+(f)\to \operatorname {Spec(k)}$ dual to the ring morphism $k\to k[T_0,\cdots ,T_n]_{(f)}$ (beware that $k[T_0,\cdots ,T_n]_{(f)}$ is the degree zero component of the $\mathbb Z$-graded ring $k[T_0,\cdots ,T_n]_f$).
The open subsets $D_+(f)$ for varying $f$ cover $\mathbb P^n_k$ and the morphisms $D_+(f)\to \operatorname {Spec(k)}$ are mutually compatible, glueing to the required morphism $\: \mathbb P^n_k \to \operatorname {Spec(k)}$
[We may restrict ourselves to the subcovering $(D_{+}(X_i))_{0\leq i\leq n}$ of $\mathbb P^n_k$ obtained by only taking the $n+1$ homogeneous polynomials $f=T_i$, at the cost of being slightly less canonical].
Remark
The Proj construction although quite powerful is difficult and full of traps.
The present question attests to the subtlety of this construction, all the more so that many textbooks give it a very crisp treatment and for example don't even mention the canonical morphism investigated here.