The rational map (x,y) mapsto y / x on the affine plane is undefined along x = 0

Not every rational map on affine space extends to a regular map everywhere.

On ${\mathbf{A}}_k^2$ with coordinates $(x,y)$, the quotient $y\mathrm{/}x$ is regular on the principal open $D(x)$ and therefore defines a rational map $\eta :{\mathbf{A}}_k^2⇢{\mathbf{A}}_k^1\mathrm{.}$ Its domain of definition contains every point with $x\mathrm{\ne }0$.

It does not extend regularly across any point $(0,b)$ with $b\mathrm{\ne }0$. Indeed, an extension near such a point would give an open neighbourhood $W$ of $(0,b)$ and a regular function $g$ on $W$ that agrees with $y\mathrm{/}x$ on $W\cap D(x)$. By Principal opens form a basis for the Zariski topology on an affine variety, choose a principal open $D(h)\subseteq W$ with $(0,b)\in D(h)$. Then Regular functions on a principal open are the principal localization of the coordinate ring identifies $g{\mathrm{\mid }}_{D(h)}$ with some fraction $a\mathrm{/}{h}^n\in k[x,y{]}_h$, and on the smaller principal open $D(h)\cap D(x)=D(hx)$ the same theorem identifies the equality $g=y\mathrm{/}x$ with the localization identity $\frac{a}{h^n}=\frac{y}{x}\text{in }k[x,y{]}_{hx}\mathrm{.}$ Thus some power of $hx$ annihilates $xa-y{h}^n$. Since $k[x,y]$ is a domain, it follows that $xa=y{h}^n$ in $k[x,y]$. Evaluating at $(0,b)$ gives $0=bh(0,b{)}^n,$ which is impossible because $b\mathrm{\ne }0$ and $(0,b)\in D(h)$ means $h(0,b)\mathrm{\ne }0$. This contradiction shows that no regular extension exists near $(0,b)$.