If $\pi$ is a prime of a UFD $A$, is $\text{Spec }A$ a coproduct of $\text{Spec }A[\pi^{-1}]$ and $\text{Spec }A_{(\pi)}$ over $\text{Spec Frac }A$?

Question

Let $A$ be a UFD (unique factorization domain) with fraction field $K$. Let $\pi\in A$ a prime. Let $A_{(\pi)}$ be the localization at the ideal $\pi$, and let $A[\pi^{-1}]$ be the localization w.r.t. the multiplicative subset $\{1,\pi,\pi^2,\ldots\}$.

Because $A$ is a UFD, one checks that $A[\pi^{-1}]\cap A_{(\pi)} = A$, the intersection taken inside the fraction field $K$. This implies that $A$ is ring-theoretic fiber product of $A[\pi^{-1}]$ and $A_{(\pi)}$ over $K$, and hence $\text{Spec }A$ is the affine scheme coproduct of $\text{Spec }A[\pi^{-1}]$ and $\text{Spec }A_{(\pi)}$ over $\text{Spec }K$.

Is $\text{Spec }A$ also a scheme coproduct?

Answer 1

No. Take $A=\mathbb C[x,y]$ and $\pi=x$. Consider the rational map $f:\mathbf A^2=\mathrm{Spec}A\to\mathbf P^1:(x,y)\mapsto [x:y]$. Then $f$ is a morphism on both $\mathrm{Spec}A_{(\pi)}$ and $A[\pi^{-1}]$ since $\frac xy\in A_{(\pi)}$ and $\frac yx\in A[\pi^{-1}]$, respectively. But $f$ itself is not defined in the point $(0,0)$.