Resolving the rational map [x:y] at the origin

The two coordinate functions $x,y$ are global sections of the line bundle ${\mathcal{O}}_{{\mathbb{A}}^2}$; they generate it over ${\mathbb{A}}^2\setminus \{0\}=D(x)\cup D(y)$, and the image of the map ${\mathcal{O}}_{{\mathbb{A}}^2}^2\to {\mathcal{O}}_{{\mathbb{A}}^2}$, $(a,b)\mapsto ax+by$, is the ideal $(x,y)$. Hence $\phi$ is the rational map attached by [F3] to this pair of sections, and its base ideal is $I=(x,y)$, the maximal ideal of the origin.

By [F1] applied to $X={\mathbb{A}}_k^2$, $\mathcal{L}=\mathcal{O}$ and the sections $x,y$, the blowup ${\mathrm{Bl}}_I{\mathbb{A}}_k^2$ resolves $\phi$: the induced map is the unique morphism extending $\phi$, characterized by the pulled-back sections. Since $I=(x,y)$ is the ideal of the origin, ${\mathrm{Bl}}_I{\mathbb{A}}_k^2={\mathrm{Bl}}_0{\mathbb{A}}_k^2$, which by [F2] is the incidence subscheme $Z=V(xv-yu)\subseteq {\mathbb{A}}_k^2\times {\mathbb{P}}_k^1$; on the chart with coordinates $(x,s)$, $y=xs$, one has $[x:y]=[1:s]$ and the second projection sends a point to $[1:s]$, while on the chart with coordinates $(t,y)$, $x=yt$, it sends a point to $[t:1]$. Both formulas agree with $(x,y)\mapsto [x:y]$ wherever the latter is defined, so the second projection is the resolved morphism.

The exceptional curve of the blowup is $E={\pi }^{-1}(0)$, the fibre of the first projection of $Z$ over the origin. Over $0$ the conditions $x=y=0$ become vacuous in the incidence equation, so $E=\{0\}\times {\mathbb{P}}_k^1$ and the second projection restricts to an isomorphism $E\to {\mathbb{P}}_k^1$; every normal direction occurs exactly once. Therefore the rational map is resolved by the blowup, the extension is the projection of the incidence model, and the exceptional curve maps isomorphically onto ${\mathbb{P}}_k^1$.