I'm going to prove following exercise from Bosch's "Algebraic Geometry and Commutative Algebra" (page 461):
I'm struggle with proving that if $m < n$ then the morphism $\varphi: \mathbb{P}^n_R \to \mathbb{P}^m_R$ must be constant.
I tried following:
Assume that we have such non constant $\varphi$. Then we get $m+1$ global sections $s_i \in \Gamma(\mathbb{P}^n, \mathcal{L})$ where $\mathcal{L}:= \varphi^*\mathcal{O}_{P^m}(1)$ where $s_i:= f^{\#}(t_i)$ for canonical $t_i \in \Gamma(\mathbb{P}^m, \mathcal{O}_{P^m}(1))= \oplus_{i=0}^m \mathcal{O}_{P^m}(0) \cdot t_i$.
Since $Pic(\mathbb{P}^n)= \mathbb{Z}$ we conclude $f^*\mathcal{O}_{P^m}(1)= \mathcal{O}_{P^n}(d)$ for a $d \ge 0$ (otherwise it would not have global sections.
Denote $U:= span(s_i \vert i \in \{0,1,...,m\})= <f^{\#}(t_i) \vert i \in \{0,1,...,m\}>_R $ the $R$-module subspace of $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$.
Since the $t_i$ generate $\Gamma(\mathbb{P}^m, \mathcal{O}_{P^m}(1))$ then the pullbacks $s_i$ generate $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$, therefore $U= \Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d))$.
Since $m<n$ by assumption and $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(d)))$ is generated by exactly $n+1$ elements if $d>0$ and otherwise $\Gamma(\mathbb{P}^n, \mathcal{O}_{P^n}(0))=R$ is generated by one element, we conclude $d=0$.
So $\varphi^*\mathcal{O}_{P^m}(1) = \mathcal{O}_{P^n}$ is the structure sheaf and all it's global sections are constants. Therefore also $s_i$.
What I don't understand is why does it already imply that $\varphi$ is constant, so that $\varphi(\mathbb{P}^n)=pt$?
We consider now the construction for $\varphi: X \to P^m$ in detail. As in Hartshorne described it comes by glueing the restrictions $\phi_i: X_{s_i} \to D_+(t_i)$ via ring maps $k[t_0/t_i, ..., t_m/t_i] \to \Gamma(\mathcal{O}_X,X_{s_i}), t_j//t_i \mapsto s_j/s_i$. Since $s_i$ constants we get $X= X_{s_i}$ and therefore all $\varphi_i$ factorise over $Spec(\Gamma(\mathcal{O}_X,X)$. Then also $\varphi$ factorise through this affine scheme.
So that it indeed suffice to show that the composition $X_{s_i}\to Spec(\Gamma(\mathcal{O}_X,X) \to D_+(t_i)$ is already constant. But how?
In general, suppose $\phi:X\to \mathbb{P}^n$ is a morphism. This is equivalent to the data of a line bundle $L$ on X and $n+1$ sections $s_0,\ldots,s_n$ that generates $L$ locally just like you said.
Conversely, given a line bundle $L$ over $X$ and $n+1$ section $s_0,\ldots , s_n$ on $X$, how can we build a morphism to $\mathbb{P}^n$ ?
Fix $x\in X$. Then $s_0(x),\ldots,s_n(x)$ can't all be zero in $L_x$.
So the zeroes of $(\lambda_0,...,\lambda_n)\mapsto \lambda_0 s_0(x)+\ldots\lambda_n s_n(x)$ defines a hyperplane in $K^n$ or, by duality, a line $l_x$ in $(K^n)^*$ -- that is a point in the projective $\mathbb{P}^n$. Then put $\phi(x)=[l_x]$.
In your case, since all the sections are constant, the zeroes of map $(\lambda_0,...,\lambda_n)\to \lambda_0 s_0(x)+\ldots\lambda_n s_n(x)$ are independant of $x$ : so the map you get to the projective is, in the end, itself constant.