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.



Example showing that $\mathbb{P}^1$ does not preserve monics
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.


