Example showing that $\mathbb{P}^1$ does not preserve monics Is there an injective homomorphism of commutative rings $A \to B$ such that the induced map $\mathbb{P}^1(A) \to \mathbb{P}^1(B)$ is not injective? Here, $\mathbb{P}^1(A) = \mathrm{Hom}(\mathrm{Spec}(A),\mathbb{P}^1)$ is the set of equivalence classes of invertible quotients of $A^2$.
It is easy to check that the map is injective on the subset of free quotients of $A^2$, i.e. points which can be defined by homogeneous coordinates. In particular, the whole map is injective if $\mathrm{Pic}(A)$ is trivial.
Edit. Sorry now I realize that this is a rather stupid question. If $X$ is any separated scheme and $f : A \to B$ is injective, then $X(A) \to X(B)$ is injective as well.
 A: There is no such injective ring homomorphism.  For every pair of rank $1$, locally free quotient $A$-modules, $$q_i:A^{\oplus 2}\twoheadrightarrow Q_i,\ i=1,2,$$ there exists a finite set of elements of $A$, say $(a_j)_{j=1,\dots,n}$, generating the unit ideal and trivializations $$h_{i,j}:Q_i[1/a_j]\xrightarrow{\cong} A[1/a_j].$$  Consider the compositions, $$h_{i,j}\circ q_i: A[1/a_j]^{\oplus 2}\twoheadrightarrow A[1/a_j].$$  If the induced morphisms $g_{B,i}:B^{\oplus 2}\twoheadrightarrow B\otimes_A Q_i$ are equal as morphisms from $\text{Spec}(B)$ to $\mathbb{P}^1$, then there exists an elements $\beta_{(1,2),j}$ and $\beta_{(2,1),j}$ of $B[1/a_j]$ such that the composition $h_{2,j,B}$ equals the scalar product of $\beta_{(1,2),j}$ and the composition $h_{1,j,B}$, and similarly $h_{1,j,B}$ equals the scalar product of $\beta_{(2,1),j}$ and $h_{2,j,B}$.  By surjectivity of $h_{i,j,B}$, also $\beta_{(2,1),j}\cdot \beta_{(1,2),j}$ equals $1$.
Since $h_{i,j}$ is surjective, there exists an element $u_{i,j}$ that maps to $1$.  Thus, $\beta_{(1,2),j}$ equals the image in $B[1/a_j]$ of $\alpha_{(1,2),j}:=h_{2,j}(u_{1,j})$, and $\beta_{(2,1),j}$ equals the image of $\alpha_{(2,1),j}:=h_{1,j}(u_{2,j})$.
Since $A[1/a_j]$ is a flat $A$-module, also $A[1/a_j]\to B[1/a_j]$ is injective.  Since the composition $h_{2,j}$ equals the scalar multiple of $\alpha_{(1,2),j}$ with $h_{1,j}$ after base change to $B[1/a_j]$, these $A[1/a_j]$-module homomorphisms are equal.  Similarly, $h_{1,j}$ equals the scalar multiple of $\alpha_{(2,1),j}$ and $h_{2,j}$.  Since the base change to $B[1/a_j]$ of $\alpha_{(1,2),j}\cdot \alpha_{(2,1),j}$ equals the base change of $1$, also $\alpha_{(1,2),j}\cdot \alpha_{(2,1),j}$ equals $1$.  Thus, the morphisms $f_1$ and $f_2$ are equal.
